CBSE Sample Paper 2020-21 · Part B: Section IV · 3 marks

Q30: $\frac{dy}{dx}$ for $y = e^{x\sin^2 x} + (\sin x)^x$

Sample 2020-21Ch 5 · Continuity and Differentiability

The question

If $y = e^{x\sin^2 x} + (\sin x)^x$ ⁠, find $\dfrac{dy}{dx}$ ⁠.

Exam answer

Let $u = e^{x\sin^2 x}$ and $v = (\sin x)^x$ ⁠, so that $y = u + v$ ⁠.

\Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \quad\ldots(\text{i})

Why?

$(\sin x)^x$ has $x$ in both the base and the power, so no single rule differentiates it; taking logs does. But logs do not split a sum:

\log(u + v) \ne \log u + \log v

So we name the two terms $u$ and $v$ ⁠, and differentiate each one separately.

Now $u = e^{x\sin^2 x}$ ⁠. Differentiating both sides w.r.t. $x$ ⁠,

\begin{aligned} \frac{du}{dx} &= e^{x\sin^2 x}\frac{d}{dx}(x\sin^2 x) \\ &= e^{x\sin^2 x}[x(2\sin x\cos x) \\ &\qquad + \sin^2 x] \\ &= e^{x\sin^2 x}[x\sin 2x \\ &\qquad + \sin^2 x] \quad\ldots(\text{ii}) \end{aligned}

Why?

Chain rule: $\frac{d}{dx}e^{g(x)} = e^{g(x)}\,g'(x)$ ⁠, with $g(x) = x\sin^2 x$ ⁠.

Product rule for $g$ ⁠, using $\frac{d}{dx}\sin^2 x = 2\sin x\cos x$ ⁠:

Write $g = pq$ with the two parts and their derivatives:

\begin{aligned} p &= x, & p' &= 1 \\ q &= \sin^2 x, & q' &= 2\sin x\cos x \end{aligned}

For $q'$ ⁠, $\sin^2 x = (\sin x)^2$ ⁠: the outside is a square, the inside is $\sin x$ ⁠. So $q' = 2(\sin x)^1 \cdot \cos x$ ⁠.

Then $g' = pq' + qp'$ ⁠:

\begin{aligned} g'(x) &= x(2\sin x\cos x) \\ &\quad + \sin^2 x \cdot 1 \end{aligned}

Then $2\sin x\cos x = \sin 2x$ ⁠.

Also $v = (\sin x)^x$ ⁠, so

\log v = x\log(\sin x)

Why?

Take logs of both sides and use $\log(a^b) = b\log a$ ⁠, which brings the power $x$ down. This needs $\sin x > 0$ ⁠, which is where $(\sin x)^x$ is defined.

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

\begin{aligned} \frac1v\frac{dv}{dx} &= x\cdot\frac{\cos x}{\sin x} + \log(\sin x) \\ &= x\cot x + \log(\sin x) \end{aligned}

Why?

Left side, by the chain rule: $\frac{d}{dx}\log v = \frac1v\frac{dv}{dx}$ ⁠.

Right side, by the product rule, with

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

and $\frac{\cos x}{\sin x} = \cot x$ ⁠.

\begin{aligned} p &= x, & p' &= 1 \\ q &= \log(\sin x), & q' &= \frac{1}{\sin x}\cdot\cos x \end{aligned}

\begin{aligned} pq' + qp' &= x \cdot \frac{\cos x}{\sin x} \\ &\quad + \log(\sin x) \cdot 1 \end{aligned}

\begin{aligned} \frac{dv}{dx} &= (\sin x)^x[x\cot x \\ &\qquad + \log(\sin x)] \quad\ldots(\text{iii}) \end{aligned}

Why?

Multiply both sides by $v$ ⁠, and put back $v = (\sin x)^x$ ⁠. Watch out: the answer must not contain $v$ ⁠.

\begin{aligned} \frac{dv}{dx} &= v\,[x\cot x + \log(\sin x)] \\ &= (\sin x)^x[x\cot x \\ &\qquad + \log(\sin x)] \end{aligned}

Substituting (ii) and (iii) in (i),

$\dfrac{dy}{dx} = e^{x\sin^2 x}[x\sin 2x + \sin^2 x]$ $+ (\sin x)^x[x\cot x$ $+ \log(\sin x)]$

Where this comes from

Chapter 5, Continuity and Differentiability: Logarithmic differentiation of a sum.

Question type: Logarithmic differentiation

How to spot it: click to see

Type 7. Logarithmic differentiation: $x$ in the power

Spot it: a power whose exponent contains $x$ ⁠, like $x^x$ ⁠, $x^{\sin x}$ or $(\log x)^{\cos x}$ ⁠, or a long product.

Here: $x$ is in the power of $\sin x$ in $(\sin x)^x$ ⁠: name $u = e^{x\sin^2x}$ and $v = (\sin x)^x$ ⁠. Do $u$ by the chain rule, and take log of $v$ to get $x\log(\sin x)$ ⁠.

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Q30: $\frac{dy}{dx}$ for $y = e^{x\sin^2 x} + (\sin x)^x$