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CBSE Board Examination 2024 · Section B · 2 marks

Q21: Is f(x)=x2∣x∣f(x) = x^2|x| differentiable at x=0x = 0⁠? or prove x dydx=1+y44y\sqrt x\,\frac{dy}{dx} = \frac{1 + y^4}{4y} for y=tan⁡xy = \sqrt{\tan\sqrt x}

Board 2024Ch 5 · Continuity and Differentiability
The question

(a) Check whether the function f(x)=x2 ∣x∣f(x) = x^2\,|x| is differentiable at x=0x = 0 or not.

OR

(b) If y=tan⁡xy = \sqrt{\tan\sqrt x}⁠, prove that x dydx=1+y44y\sqrt x\,\dfrac{dy}{dx} = \dfrac{1 + y^4}{4y}⁠.

Exam answer

(a)

f(x)=x2∣x∣f(x) = x^2|x|⁠. As ∣x∣=x|x| = x for x≥0x \ge 0 and ∣x∣=−x|x| = -x for x<0x < 0⁠,

f(x)={x3,x≥0−x3,x<0f(x) = \begin{cases} x^3, & x \ge 0 \\ -x^3, & x < 0 \end{cases}
Why?

A modulus is removed by splitting the real line where the expression inside it, here xx⁠, changes sign:

∣x∣={x,x≥0−x,x<0|x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}

Multiplying each piece by x2x^2 gives x3x^3 and −x3-x^3⁠. The point x=0x = 0 is exactly where the formula changes, so the derivative there must be found from each side separately.

RHD=lim⁡h→0f(0+h)−f(0)h=lim⁡h→0h3h=lim⁡h→0h2=0\begin{aligned} \text{RHD} &= \lim_{h \to 0}\frac{f(0 + h) - f(0)}{h} \\ &= \lim_{h \to 0}\frac{h^3}{h} = \lim_{h \to 0}h^2 = 0 \end{aligned}
Why?

f is differentiable at a point when the right-hand derivative (RHD) and the left-hand derivative (LHD) there exist and are equal.

For the RHD, h>0h > 0⁠, so 0+h0 + h is on the right of 0 and we use the piece x3x^3⁠: f(h)=h3f(h) = h^3⁠. Also f(0)=0f(0) = 0⁠.

f(0)=(0)3=0f(0+h)=f(h)=h3f(h)−f(0)h=h3−0h=h3h=h2(cancel h)lim⁡h→0h2=(0)2=0\begin{aligned} f(0) &= (0)^3 = 0 \\ f(0 + h) &= f(h) = h^3 \\[4pt] \frac{f(h) - f(0)}{h} &= \frac{h^3 - 0}{h} \\ &= \frac{h^3}{h} = h^2 \quad (\text{cancel } h) \\[4pt] \lim_{h \to 0} h^2 &= (0)^2 = 0 \end{aligned}
LHD=lim⁡h→0f(0−h)−f(0)−h=lim⁡h→0h3−h=lim⁡h→0(−h2)=0\begin{aligned} \text{LHD} &= \lim_{h \to 0}\frac{f(0 - h) - f(0)}{-h} \\ &= \lim_{h \to 0}\frac{h^3}{-h} = \lim_{h \to 0}(-h^2) = 0 \end{aligned}
Why?

Now 0−h<00 - h < 0⁠, so we use the piece −x3-x^3⁠:

f(−h)=−(−h)3=−(−h3)=h3f(−h)−f(0)−h=h3−0−h=−h3h=−h2lim⁡h→0(−h2)=−(0)2=0\begin{aligned} f(-h) &= -(-h)^3 \\ &= -(-h^3) \\ &= h^3 \\[4pt] \frac{f(-h) - f(0)}{-h} &= \frac{h^3 - 0}{-h} \\ &= -\frac{h^3}{h} = -h^2 \\[4pt] \lim_{h \to 0} (-h^2) &= -(0)^2 = 0 \end{aligned}

((−h)3=(−h)(−h)(−h)=−h3(-h)^3 = (-h)(-h)(-h) = -h^3⁠, since three minus signs multiply to a minus.)

Watch out: the denominator for the LHD is −h-h⁠, the change in xx when we step from 0 to 0−h0 - h⁠.

RHD = LHD = 0, so f(x)f(x) is differentiable at x=0x = 0⁠.

(b)

y=tan⁡xy = \sqrt{\tan\sqrt x}⁠. Differentiating w.r.t. xx (chain rule),

Why?

y is built in three layers: a square root, of a tan, of a square root. The chain rule differentiates the outside layer first and multiplies by the derivative of what is inside, layer by layer:

ddxu=12u⋅dudx\frac{d}{dx}\sqrt u = \frac{1}{2\sqrt u}\cdot\frac{du}{dx} ddxtan⁡u=sec⁡2u⋅dudx\frac{d}{dx}\tan u = \sec^2u\cdot\frac{du}{dx}
dydx=sec⁡2x2tan⁡x×12x\frac{dy}{dx} = \frac{\sec^2\sqrt x}{2\sqrt{\tan\sqrt x}}\times\frac{1}{2\sqrt x}
Why?

The outside root (with u=tan⁡xu = \tan\sqrt x⁠) gives 12tan⁡x\frac{1}{2\sqrt{\tan\sqrt x}}⁠. The tan layer gives sec⁡2x\sec^2\sqrt x⁠. The inside root gives 12x\frac{1}{2\sqrt x}⁠.

x dydx=sec⁡2x4tan⁡x=1+tan⁡2x4tan⁡x=1+(tan⁡x)44tan⁡x\begin{aligned} \sqrt x\,\frac{dy}{dx} &= \frac{\sec^2\sqrt x}{4\sqrt{\tan\sqrt x}} \\ &= \frac{1 + \tan^2\sqrt x}{4\sqrt{\tan\sqrt x}} \\ &= \frac{1 + \left(\sqrt{\tan\sqrt x}\right)^4}{4\sqrt{\tan\sqrt x}} \end{aligned}
Why?

Multiplying by x\sqrt x cancels the x\sqrt x in the denominator, and 2×2=42 \times 2 = 4⁠.

x×sec⁡2x2tan⁡x×12x=x sec⁡2x2×2x tan⁡x=sec⁡2x4tan⁡x(÷x)\begin{aligned} &\sqrt x \times \frac{\sec^2\sqrt x}{2\sqrt{\tan\sqrt x}} \times \frac{1}{2\sqrt x} \\ &= \frac{\sqrt x\,\sec^2\sqrt x}{2 \times 2\sqrt x\,\sqrt{\tan\sqrt x}} \\ &= \frac{\sec^2\sqrt x}{4\sqrt{\tan\sqrt x}} \quad (\div \sqrt x) \end{aligned}

Then use sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta with θ=x\theta = \sqrt x⁠.

Finally, to bring in y, write tan⁡x=(tan⁡x)2=y2\tan\sqrt x = \left(\sqrt{\tan\sqrt x}\right)^2 = y^2⁠. So tan⁡2x=y4\tan^2\sqrt x = y^4 and tan⁡x=y\sqrt{\tan\sqrt x} = y⁠.

y=tan⁡xy2=tan⁡x(square)y4=(y2)2=tan⁡2x1+tan⁡2x4tan⁡x=1+y44y\begin{aligned} y &= \sqrt{\tan\sqrt x} \\ y^2 &= \tan\sqrt x \quad (\text{square}) \\ y^4 &= (y^2)^2 = \tan^2\sqrt x \\[4pt] \frac{1 + \tan^2\sqrt x}{4\sqrt{\tan\sqrt x}} &= \frac{1 + y^4}{4y} \end{aligned}

∴ x dydx=1+y44y\therefore\ \sqrt x\,\dfrac{dy}{dx} = \dfrac{1 + y^4}{4y}⁠. Hence proved.

Where this comes from

Chapter 5, Continuity and Differentiability: Differentiability at a point; chain rule.

Question type: (a) Is ff differentiable at a point? Compare LHD and RHD; (b) The chain rule

How to spot it: click to see

(a) Type 3. Is ff differentiable at a point? Compare LHD and RHD

Spot it: a modulus such as ∣x−1∣|x - 1|⁠, or a function in pieces, and the word "differentiable".

Here: ∣x∣|x| inside x2∣x∣x^2|x|⁠, and the word differentiable: write ff as x3x^3 for x≥0x \ge 0 and −x3-x^3 for negative xx⁠. Both LHD and RHD at 0 come out 0, so it is differentiable.

(b) Type 4. The chain rule: a function inside a function

Spot it: something other than a plain xx inside sin⁡\sin⁠, cos⁡\cos⁠, log⁡\log⁠, ee or a power, like sin⁡(x2+5)\sin(x^2 + 5) or ecos⁡xe^{\cos x}⁠.

Here: Three layers: a root, of tan⁡\tan⁠, of x\sqrt x⁠. Differentiate from the outside in to get sec⁡2x2tan⁡x⋅12x\frac{\sec^2\sqrt x}{2\sqrt{\tan\sqrt x}}\cdot\frac{1}{2\sqrt x}⁠, then write sec⁡2=1+tan⁡2\sec^2 = 1 + \tan^2 in terms of yy⁠.

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