If y-cos x3- then find dydx at x-0

We have, y=cos x^3 Let u=x^3, Differentiating u w.r.t x, we get dudx=3x^2 Also, y=cos u Differentiating w.r.t u, dydu= sin u As we know by chain rule, dydx=dydu ...

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Standard XII

Physics

Chain Rule of Differentiation

If y=cos x3, ...

Question

If $y = cos x^{3}$ , then find $\frac{d y}{d x}$ at $x = 0$

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Solution

The correct option is A $0$
We have, $y = cos x^{3}$
Let $u = x^{3}$ ,
Differentiating $u$ w.r.t $x$ , we get
$\frac{d u}{d x} = 3 x^{2}$
Also, $y = cos u$
Differentiating w.r.t $u$ ,
$\frac{d y}{d u} = - sin u$
As we know by chain rule,
$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$
$\frac{d y}{d x} = - sin u \cdot 3 x^{2}$
$\frac{d y}{d x} = - (sin x^{3}) \cdot 3 x^{2}$
${∣ ∣ ∣ \frac{d y}{d x} ∣ ∣ ∣}_{x = 0} = (sin 0) \cdot 3 \times 0 = 0$
So, option A is correct.

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If y=cosx3, then find dydx at x=0

The correct option is A 0We have, y=cos x3 Let u=x3, Differentiating u w.r.t x, we get dudx=3x2 Also, y=cosu Differentiating w.r.t u, dydu=−sinu As we know by chain rule, dydx=dydu⋅dudx dydx=−sinu⋅3x2 dydx=−(sinx3)⋅3x2 ∣∣∣dydx∣∣∣x=0=(sin0)⋅3×0=0 So, option A is correct.

If $y = cos x^{3}$ , then find $\frac{d y}{d x}$ at $x = 0$