Differentiation

We have to find the derivative of e^–x with the first principle method, so,

f(x) = e^–x

by using the first principle formula, we get,

f ‘(x) =

f ‘(x) =

f ‘(x) =

f ‘(x) =

[By using = 1]

f ‘(x) = – e^–x

We have to find the derivative of e^3x with the first principle method, so,

f(x) = e^3x

by using the first principle formula, we get,

f ‘(x) =

f ‘(x) =

f ‘(x) =

f ‘(x) =

[By using = 1]

f ‘(x) = 3e^3x

We have to find the derivative of e^ax+b with the first principle method, so,

f(x) = e^ax+b

by using the first principle formula, we get,

f ‘(x) =

f ‘(x) =

f ‘(x) =

[By using = 1]

f ‘(x) = a e^ax+b

We have to find the derivative of e^cosx with the first principle method, so,

f(x) = e^cosx

by using the first principle formula, we get,

f ‘(x) =

f ‘(x) =

f ‘(x) =

f ‘(x) =

[By using = 1]

f ‘(x) =

f ‘(x) =

[By using cos(x+h) = cosx cosh – sinx sinh]

f ‘(x) =

[By using lim_x→0 = 1 and

cos 2x = 1–2sin² x]

f ‘(x) =

f ‘(x) =

f ‘(x) = –e^{cos x} sin x

We have to find the derivative of e^√2x with the first principle method, so,

f(x) = e^√2x

by using the first principle formula, we get,

f ‘(x) =

f ‘(x) =

f ‘(x) =

f ‘(x) =

[By using = 1]

f ‘(x) =

[By rationalising]

f ‘(x) =

f ‘(x) =

We have to find the derivative of log cosx with the first principle method, so,

f(x) = log cos x

by using the first principle formula, we get,

f ‘(x) =

f ‘(x) =

f ‘(x) =

f ‘(x) =

[Adding and subtracting 1]

f ‘(x) =

[Rationalising]

f ‘(x) =

[By using lim_x→0 = 1]

f ‘(x) =

[cosC – cosD = –2 sin sin]

f ‘(x) = [By using lim_x→0 = 1]

f ‘(x) =

f ‘(x) = – tan x

We have to find the derivative of with the first principle method, so,

f(x) =

by using the first principle formula, we get,

f ‘(x) =

f ‘(x) =

f ‘(x) =

f ‘(x) =

[By using lim_x→0 = 1]

f ‘(x) =

[Rationalizing]

f ‘(x) =

f ‘(x) =

[sinA cosB – cosA sinB = sin(A–B)]

f ‘(x) =

[By using lim_x→0 = 1]

f ‘(x) =

f ‘(x) =

We have to find the derivative of x²e^x with the first principle method, so,

f(x) = x²e^x

by using the first principle formula, we get,

f ‘(x) =

f ‘(x) =

f ‘(x) =

[By using (a+b)² = a²+b²+2ab]

f ‘(x) =

f ‘(x) =

[By using lim_x→0 = 1]

f ‘(x) = x²e^x + lim_h→0 e^(x+h) [h+2x]

f ‘(x) = x²e^x + 2x e^x

We have to find the derivative of log cosec x with the first principle method, so,

f(x) = log cosecx

by using the first principle formula, we get,

f ‘(x) =

f ‘(x) =

f ‘(x) =

f ‘(x) =

[By using log a – log b = log ]

f ‘(x) =

[adding and subtracting 1]

f ‘(x) =

f ‘(x) =

[Rationalising]

f ‘(x) =

f ‘(x) =

[sin C – sin D = 2 sin cos ]

f ‘(x) =

[By using lim_x→0 = 1]

f ‘(x) = – cot x

We have to find the derivative of sin^–1(2x+3) with the first principle method, so,

f(x) = sin^–1(2x+3)

by using the first principle formula, we get,

f ‘(x) =

f ‘(x) =

Let sin^–1[2(x+h)+3] = A and sin^–1(2x+3) = B, so,

sinA = [2(x+h)+3] and sinB = (2x+3),

2h = sinA – sinB, when h→0 then sinA→sinB we can also say that A→B and hence A–B→0,

f ‘(x) =

f ‘(x) =

[sinC – sinD = 2 sin cos ]

f ‘(x) =

[By using lim_x→0 = 1]

f ‘(x) =

f ‘(x) =

[By using Pythagoras theorem, in which H = 1 and P = 2x+3, so, we have to find B, which comes out to be by the relation H² = P² + B²]

f ‘(x) =

f(x) = log_e(log_ex)

Using the Chain Rule of Differentiation,

So,

(Ans)

f(x) = x + 1

(fof)(x) = f(x) + 1

= (x + 1) + 1

= x + 2

So,

=1 (Ans)

y = f(log_ex)

Using the Chain Rule of Differentiation,

So, at x = e

(Ans)

Using the Chain Rule of Differentiation, derivative of log(f(e^x)) w.r.t. x is

So, the value of the derivative at x = 0 is

So, the value of the derivative at x = 0 is 0.5 (Ans)

y = f(x²)

Putting x = 1,

=2√1

=2

i.e., at x = 1. (Ans)

From the definition of invertible function,

g(f(x)) = x …(i)

So, g(f(3)) = 3, i.e., g(9) = 3

Now, differentiating both sides of equation (i) w.r.t. x using the Chain Rule of Differentiation, we get –

g’(f(x)). f’(x) = 1 …(ii)

Plugging in x = 3 in equation (ii) gives us –

g’(f(3)).f’(3) = 1

or, g’(9).9 = 1

i.e., g’(9) = 1/9 (Ans)

For x,

=x

So, (Ans)

For ,

y=sin^-1 (sin x)

= sin^-1 ( sin (π –(π -x))

(to get y in principal range of sin^-1 x)

i.e.,

y = π - x

From the last problem we see that and

So, y is not differentiable at

Extending this, we can say that y is not differentiable at x = (2n+1)

So, for

(Ans)

y = cos^-1 (cos x)

for x (π, 2π)

y = cos^-1(cos x)

= cos^-1(cos (π + (x - π)))

= cos^-1(-cos (x-π))

= π – (x - π)

= 2π - x

[Since, cos(π+x) = - cos x and cos^-1(-x) = π-x]

So,

For cos^-1(cos x), x = n are the ‘sharp corners’ where slope changes from 1 to -1 or vice versa, i.e., the points where the curves are not differentiable.

So, for x [π,2π]

(Ans)

So,

So, answer is (Ans)

Using the Chain Rule of Differentiation,

= f(e^x). e^f(x) f^’(x) + f^’(e^x)e^x. e^f(x)

At x = 0,

= f(1). e^f(0) f’(0) + f’(1). e^f(0)

= 0. e⁰ f’(0) + 2.e⁰

= 0 + 2.1

= 2

y = x|x|

or,

So, for x < 0

=-2x (Ans)

We know that

So, here y = sin^–1 x + cos^–1 x

which is a constant.

Also, sin^-1 x and cos^-1 x exist only when -1 x 1

So, when x [-1, 1] and does not exist for all other values of x.

and

Using Chain Rule of Differentiation,

=cot θ-cosec θ (Ans)

When , tan x is negative. So, square root of tan² x in this condition is –tan x.

So,

=tan^-1 (-tan x)

=-tan^-1 (tan x)

=-x

And so

(Ans)

y = x^x

Taking logarithm on both sides,

log y = x log x

Differentiating w.r.t. x on both sides,

=1+log x

=x^x (1+log x)

So, at x = e,

=e^e (1+1)

=2e^e (Ans)

Using the Chain Rule of Differentiation,

(Ans)

y = log_a x =

(Ans)

This particular problem is a perfect way to demonstrate how simple but powerful the Chain Rule of Differentiation is.

It is important to identify and break the problem into the individual functions with respect to which successive differentiation shall be done.

In this case, this is the way to break down the problem –

i.e.,

(Ans)

holds for all .

So, , for all

( )

Hence, for all .

Which exists for and is equal to

Now,

⟹x+1>0

⟹x>-1 …(i)

Also,

⟹x≥0 or x<-1 …(ii)

Comparing equations (i) and (ii), we understand that the condition satisfying both inequalities is .

So, for x≥0,

, which is a constant

So, (Ans)

Since |x| < 1,

y = 1 + x + x² + … to ∞

(Ans)

and

We know,

Using the chain rule of differentiation,

Using Chain Rule of Differentiation,

Dividing numerator and denominator by (1+x²)²,

=sec u (1+tan u ) (Ans)

Using the Chain Rule of Differentiation,

Putting x = 1,

Since, u(1) = v(1),

2v(1) – 2u(1) = 0

i.e., f’(1) = 0 (Ans)

y = log |3x|

So,

i.e.,; (Ans)

f(x) is an even function.

This means that f(-x) = f(x).

If we differentiate this equation on both sides w.r.t. x, we get –

f’(-x).(-1) = f’(x)

or, -f’(-x) = f’(x)

i.e., f’(x) is an odd function. (Ans)

f(x) is an odd function.

This means that f(-x) = -f(x).

If we differentiate this equation on both sides w.r.t. x, we get –

f’(-x).(-1) = -f’(x)

or, f’(-x) = f’(x)

i.e., f’(x) is an even function. (Ans)

We have to find

So, we use the Chain Rule of Differentiation to evaluate this.

=-cot x (Ans)

Let y = sin(3x + 5)

On differentiating y with respect to x, we get

We know

[using chain rule]

However, and derivative of a constant is 0.

Thus,

Let y = tan²x

On differentiating y with respect to x, we get

We know

[using chain rule]

However,

Thus,

Let y = tan(x° + 45°)

First, we will convert the angle from degrees to radians.

We have

On differentiating y with respect to x, we get

We know

[using chain rule]

However, and derivative of a constant is 0.

Thus,

Let y = sin(log x)

On differentiating y with respect to x, we get

We know

[using chain rule]

However,

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

[using chain rule]

However,

Thus,

Let y = e^{tan x}

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

Thus,

Let y = sin²(2x + 1)

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

[using chain rule]

[∵ sin(2θ) = 2sinθcosθ]

However, and derivative of a constant is 0.

Thus,

Let y = log₇(2x – 3)

Recall that.

On differentiating y with respect to x, we get

We know

[using chain rule]

However, and derivative of a constant is 0.

Thus,

Let y = tan(5x°)

First, we will convert the angle from degrees to radians.

We have

On differentiating y with respect to x, we get

We know

[using chain rule]

However,

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

Thus,

Let y = log_x3

Recall that.

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We have and

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

Recall that (quotient rule)

However, and derivative of a constant is 0.

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

Recall that (uv)’ = vu’ + uv’ (product rule)

We have and

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

Recall that (quotient rule)

We know and derivative of a constant is 0.

(∵ sin²θ + cos²θ = 1)

(∵ sin²θ + cos²θ = 1)

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

Recall that (quotient rule)

However, and derivative of a constant is 0.

Thus,

Let y = (log sin x)²

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

[using chain rule]

However,

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

Recall that (quotient rule)

However, and derivative of a constant is 0.

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

Recall that (quotient rule)

However, and derivative of a constant is 0.

Thus,

Let y = e^3xcos(2x)

On differentiating y with respect to x, we get

Recall that (uv)’ = vu’ + uv’ (product rule)

We know and

[chain rule]

We have

Thus,

Let y = sin(log sin x)

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

[using chain rule]

However,

Thus,

Let y = e^{tan 3x}

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

[using chain rule]

However,

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

[using chain rule]

However,

Thus,

Let

We have sin2θ = 2sinθcosθ and 1 + cos2θ = 2cos²θ.

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

However,

[∵ sin2θ = 2sinθcosθ]

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We know

[using chain rule]

Recall that (quotient rule)

We know and derivative of a constant is 0.

(∵ sin²θ + cos²θ = 1)

Thus,

Let y = tan(e^{sin x})

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

[using chain rule]

However,

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We know and

[using chain rule]

However, and derivative of a constant is 0.

Thus,

Let

On differentiating y with respect to x, we get

Recall that (quotient rule)

We have (uv)’ = vu’ + uv’ (product rule)

We know, and

Thus,

Let y = log(cosec x – cot x)

On differentiating y with respect to x, we get

We know

[using chain rule]

We know and

Thus,

Let

On differentiating y with respect to x, we get

Recall that (quotient rule)

We know

However,

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

Recall that (quotient rule)

We know, and derivative of constant is 0.

Thus,

Let y = tan^–1(e^x)

On differentiating y with respect to x, we get

We know

[using chain rule]

However,

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

[using chain rule]

However,

Thus,

Let y = sin(2sin^–1x)

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

[using chain rule]

However,

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

[using chain rule]

However,

Thus,

Let y = log(tan^–1x)

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

Thus,

Let

On differentiating y with respect to x, we get

Recall that (quotient rule)

We have (uv)’ = vu’ + uv’ (product rule)

We know, and

However, and derivative of constant is 0.

Thus,

Let y = xsin(2x) + 5^x + k^k + (tan²x)³

On differentiating y with respect to x, we get

Recall that (uv)’ = vu’ + uv’ (product rule)

We know, and

Also, the derivation of a constant is 0.

We have and

Thus,

Let y = log(3x + 2) – x²log(2x – 1)

On differentiating y with respect to x, we get

Recall that (uv)’ = vu’ + uv’ (product rule)

We know and

We have and derivative of a constant is 0.

Thus,

Let

On differentiating y with respect to x, we get

Recall that (quotient rule)

We have (uv)’ = vu’ + uv’ (product rule)

We know and

However,

Thus,

Let y = sin²{log(2x + 3)}

On differentiating y with respect to x, we get

We know

[chain rule]

We have

As sin(2θ) = 2sinθcosθ, we have

We know

However, and derivative of a constant is 0.

Thus,

Let y = e^x log(sin 2x)

On differentiating y with respect to x, we get

We have (uv)’ = vu’ + uv’ (product rule)

We know and

[chain rule]

We have

However,

Thus,

Let

On differentiating y with respect to x, we get

We know

We have and derivative of a constant is 0.

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We know and

Also the derivative of a constant is 0.

However, and

Thus,

Let y = (sin^–1 x⁴)⁴

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

[using chain rule]

We have

Thus,

Let

On differentiating y with respect to x, we get

We have

[using chain rule]

Recall that (quotient rule)

We know

We have and derivative of a constant is 0.

Thus,

Let

On differentiating y with respect to x, we get

Recall that (quotient rule)

We have (uv)’ = vu’ + uv’ (product rule)

We know, and

However, and derivative of a constant is 0.

Thus,

Let y = 3e^–3xlog(1 + x)

On differentiating y with respect to x, we get

We have (uv)’ = vu’ + uv’ (product rule)

We know and

However, and derivative of a constant is 0.

Thus,

Let

On differentiating y with respect to x, we get

Recall that (quotient rule)

We know and derivative of a constant is 0.

[chain rule]

We know

Thus,

Let

On differentiating y with respect to x, we get

Recall that (quotient rule)

We have (uv)’ = vu’ + uv’ (product rule)

We know and

However, and derivative of a constant is 0.

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

However, and derivative of a constant is 0.

[∵ sin2θ = 2sinθcosθ]

[∵ sin(90°+θ) = cosθ]

Thus,

Let y = e^axsec(x)tan(2x)

On differentiating y with respect to x, we get

We have (uv)’ = vu’ + uv’ (product rule)

We will use the product rule once again.

We know, and

However,

Thus,

Let y = log(cos x²)

On differentiating y with respect to x, we get

We have

[using chain rule]

We know

[using chain rule]

However,

Thus,

Let y = cos(log x)²

On differentiating y with respect to x, we get

We have

[using chain rule]

We know

[chain rule]

However,

Thus,

Let

On differentiating y with respect to x, we get

We know

[using chain rule]

We know

[using chain rule]

Recall that (quotient rule)

We know and derivative of a constant is 0.

Thus,

Given

On differentiating y with respect to x, we get

We know

[using chain rule]

We know

However, and derivative of a constant is 0.

Thus,

Given

On differentiating y with respect to x, we get

We know

However, and derivative of a constant is 0.

But,

Thus,

Given

On differentiating y with respect to x, we get

Recall that (quotient rule)

However, and derivative of a constant is 0.

On multiplying both sides with x, we get

But,

Thus,

Given

On differentiating y with respect to x, we get

We know

[using chain rule]

We know

Thus,

Given

On differentiating y with respect to x, we get

We know

[using chain rule]

We know

[using chain rule]

Recall that (quotient rule)

We know and derivative of a constant is 0.

(∵ sec²θ – tan²θ = 1)

But, cos²θ + sin²θ = 1 and cos²θ – sin²θ = cos(2θ).

Thus,

Given

On differentiating y with respect to x, we get

We know

Thus,

Given

On differentiating y with respect to x, we get

Recall that (quotient rule)

We have (uv)’ = vu’ + uv’ (product rule)

We know and

However, and derivative of a constant is 0.

But,

Thus,

Given

On differentiating y with respect to x, we get

Recall that (quotient rule)

We know

But,

Thus,

Given y = (x – 1)log(x – 1) – (x + 1)log(x + 1)

On differentiating y with respect to x, we get

Recall that (uv)’ = vu’ + uv’ (product rule)

We know and.

Also, the derivative of a constant is 0.

Thus,

Given y = e^xcos(x)

On differentiating y with respect to x, we get

Recall that (uv)’ = vu’ + uv’ (product rule)

We know and

[chain rule]

We know

However, cos(A)cos(B) – sin(A)sin(B) = cos(A + B)

Thus,

Given

We have 1 + cos(2θ) = 2cos²θ and 1 + cos(2θ) = 2sin²θ.

[∵ log(a^m) = m × log(a)]

On differentiating y with respect to x, we get

We know

[using chain rule]

However,

We have sin(2θ) = 2sinθcosθ

Thus,

Given

On differentiating y with respect to x, we get

We have (uv)’ = vu’ + uv’ (product rule)

We know and

However, and derivative of a constant is 0.

Thus,

Given

On differentiating y with respect to x, we get

We know

[using chain rule]

However, and derivative of a constant is 0.

But,

Thus,

Given y = e^x + e^–x

On differentiating y with respect to x, we get

We know

[using chain rule]

We have

But, y = e^x + e^–x

Thus,

Given

On differentiating y with respect to x, we get

We know

[using chain rule]

However, and derivative of a constant is 0.

But,

Thus,

Given xy = 4

On differentiating y with respect to x, we get

We know

Now, we will evaluate the LHS of the given equation.

However, xy = 4

[∵ xy = 4]

Thus,

Let

On differentiating y with respect to x, we get

We have (uv)’ = vu’ + uv’ (product rule)

We know and

However, and derivative of a constant is 0.

Thus,

Now

Using sin²θ + cos²θ = 1 and 2sinθcosθ = sin2θ

= cos^–1(2cosθsinθ )

= cos^–1(sin2θ)

Considering the limits,

Therefore,

Differentiating w.r.t x,

Now

Using cos2θ = 2cos²θ – 1

y = cos^–1(cosθ)

Considering the limits,

–1< x < 1

–1< cos2θ < 1

0 < 2θ < π

Now, y = cos^–1(cosθ)

y = θ

Differentiating w.r.t x, we get

Now

Using cos2θ = 1 – 2sin²θ

y = sin^–1(sinθ)

Considering the limits,

0 < x < 1

0 < cos2θ < 1

Now, y = sin^–1(sinθ)

y = θ

Differentiating w.r.t x, we get

Now

Using sin²θ + cos²θ = 1

y = sin^–1(sinθ)

Considering the limits,

0 < x < 1

0 < cos θ < 1

Now, y = sin^–1(sinθ)

y = θ

y = cos^–1x

Differentiating w.r.t x, we get

Let x = a sinθ

Now

Using sin²θ + cos²θ = 1

y = tan^–1(tanθ)

Considering the limits,

–a < x < a

–a < asin θ < a

–1 < sin θ < 1

Now, y = tan^–1(tanθ)

y = θ

Differentiating w.r.t x, we get

Let x = a tanθ

Now

Using 1 + tan²θ = sec²θ

y = sin^–1(sinθ)

y = θ

Differentiating w.r.t x, we get

Now

Using 2cos²θ – 1 = cos2θ

y = sin^–1(cos2θ)

Considering the limits,

0 < x < 1

0 < cos θ < 1

0 < 2θ < π

0 > –2θ > –π

Now,

Differentiating w.r.t x, we get

Now

Using 1 – 2sin²θ = cos2θ

y = sin^–1(cos2θ)

Considering the limits,

0 < x < 1

0 < sin θ < 1

0 < 2θ < π

0 > –2θ > –π

Now,

Differentiating w.r.t x, we get

Let x = a cotθ

Now

Using 1 + cot²θ = cosec²θ

y = cos^–1(cosθ)

y = θ

Differentiating w.r.t x, we get

Now

Using sin(A + B) = sinA cosB + cosA sinB

Considering the limits,

Differentiating it w.r.t x,

Now

Using cos(A – B) = cosA cosB + sinA sinB

Considering the limits,

Now,

Differentiating it w.r.t x,

Let x = sinθ

Now

Using sin²θ + cos²θ = 1

Using 2cos²θ = 1 + cos2θ and 2sinθ cosθ = sin2θ

Considering the limits,

–1 < x < 1

–1 < sin θ < 1

Now,

Differentiating w.r.t x, we get

Let x = a sinθ

Now

Using sin²θ + cos²θ = 1

Using 2cos²θ = 1 + cosθ and 2sinθ cosθ = sin2θ

Considering the limits,

–a < x < a

–1 < sin θ < 1

Now,

Differentiating w.r.t x, we get

Let x = sinθ

Now

Using sin²θ + cos²θ = 1

Now

Using sin(A + B) = sinA cosB + cosA sinB

Considering the limits,

–1 < x < 1

–1 < sin θ < 1

Now,

Differentiating w.r.t x, we get

Let x = sinθ

Now

Using sin²θ + cos²θ = 1

Now

Using cos(A – B) = cosA cosB + sinA sinB

Considering the limits,

–1 < x < 1

–1 < sin θ < 1

Now,

Differentiating w.r.t x, we get

Let 2x = tanθ

Considering the limits,

–1 < 2x < 1

–1 < tanθ < 1

Now,

y = tan^–1(tan2θ)

y = 2θ

y = 2tan^–1(2x)

Differentiating w.r.t x, we get

Let 2^x = tanθ

Considering the limits,

–∞ < x < 0

2^–∞ < 2^x < 2⁰

0 < tanθ < 1

Now,

y = tan^–1(tan2θ)

y = 2θ

y = 2tan^–1(2^x)

Differentiating w.r.t x, we get

Let a^x = tanθ

Considering the limits,

–∞ < x < 0

a^–∞ < a^x < a⁰

0 < tanθ < 1

Now,

y = tan^–1(tan2θ)

y = 2θ

y = 2tan^–1(a^x)

Differentiating w.r.t x, we get

Let x = cos2θ

Now

Using 1 – 2sin²θ = cos2θ and 2cos²θ – 1 = cos2θ

Now

Using sin(A + B) = sinA cosB + cosA sinB

Considering the limits,

0 < x < 1

0 < cos 2θ < 1

Now,

Differentiating w.r.t x, we get

Let ax = tanθ