Question

if y=xcosx then findf f"(x) at x=0​

Answer 1

Question :-

• If y = f(x) = xcosx then find f"(x) at x=0.

Answer

Given :-

• f(x) = x cosx

To find :-

• f''(x) at x = 0.

Formulas used :-

$\bf\dfrac{d(x^n)}{dx}=nx^{n-1}$

$\bf\dfrac{d(x)}{dx}=1$

$\bf\dfrac{d(sinx)}{dx} = cosx$

$\bf\dfrac{d(cosx)}{dx} = - sinx$

$\bf\dfrac{d(uv)}{dx} =v \dfrac{d(u)}{dx} + u\dfrac{d(v)}{dx}$

$\bf\dfrac{df(x)}{dx} = f'(x)$

$\bf\dfrac{df'(x)}{dx} = f''(x)$

Solution :-

f(x) = x cosx

Differentiate w. r. t. x, we get

$\bf\dfrac{df(x)}{dx} = \bf\dfrac{d}{dx}xcosx$

$\bf\implies \:f'(x) = x\dfrac{d}{dx}cosx + cosx\dfrac{d}{dx}x$

$\bf\implies \:f'(x) = - xsinx + cosx$

Differentiate w. r. t. x, we get

$\bf \:\dfrac{d}{dx}f'(x) = \dfrac{d}{dx}( - xsinx + cosx)$

$\bf\implies \:f''(x) = - \dfrac{d}{dx}(xsinx) + \dfrac{d}{dx}cosx$

$\bf\implies \:f''(x) = - (x\dfrac{d}{dx}sinx + sinx\dfrac{d}{dx}x) - sinx$

$\bf\implies \:f''(x) = - (xcosx + sinx) - sinx$

$\bf\implies \:f''(x) = - xcosx - 2sinx$

Put x = 0, we get

$\bf\implies \:f''(0) = - 0 \times cos0 - 2 \times sin0$

$\bf\implies \:f''(0) = 0$

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