If y = e^{\sin^2(3x)}, what is \frac{dy}{dx}? — JAMB UTME

If $y = e^{\sin^2(3x)}$ , what is $\frac{dy}{dx}$ ?

The derivative

\frac{dy}{dx}

3\sin(6x)e^{\sin^2(3x)}

Differentiating

y = e^{\sin^2(3x)}

requires applying the chain rule step-by-step: first differentiating the exponential function to get

e^{\sin^2(3x)}

, then differentiating

\sin^2(3x)

to obtain

2\sin(3x) \cdot 3\cos(3x) = 6\sin(3x)\cos(3x)

. Multiplying these together and applying the double-angle identity

2\sin(3x)\cos(3x) = \sin(6x)

yields

3\sin(6x)e^{\sin^2(3x)}

Identify the inner function

u

and outer function

y

for applying the chain rule.

Let

u = \sin^2(3x) = (\sin(3x))^2

, so

y = e^u

The function is an exponential function whose exponent is a composite trigonometric function.

Differentiate

u = (\sin(3x))^2

with respect to

x

using the chain rule.

\frac{du}{dx} = 2\sin(3x) \cdot \frac{d}{dx}(\sin(3x)) = 2\sin(3x) \cdot 3\cos(3x) = 6\sin(3x)\cos(3x).

The derivative of

[g(x)]^2

2g(x)g'(x)

, and

\frac{d}{dx}(\sin(3x)) = 3\cos(3x)

Differentiate

y = e^u

with respect to

x

using

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

\frac{dy}{dx} = e^{\sin^2(3x)} \cdot 6\sin(3x)\cos(3x).

The derivative of

e^u

with respect to

u

e^u

Simplify the expression using the trigonometric double-angle identity

2\sin\theta\cos\theta = \sin(2\theta)

, where

\theta = 3x

\frac{dy}{dx} = 3 \cdot (2\sin(3x)\cos(3x)) e^{\sin^2(3x)} = 3\sin(6x)e^{\sin^2(3x)}.

Rewriting

6\sin(3x)\cos(3x)

3\sin(6x)

simplifies the expression to standard examination form.

Chain Rule for Composite Exponential and Trigonometric Functions

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