Question

If h(x)=2x cos x, find h″(π)​

Answer 1

Answer:

h′′(π)=6.28

Explanation:

h(x)=2x⋅cosx

h′(x)=2cosx−2xsinx (product rule)

h′′(x)=−2sinx−2sinx−2xcosx (product rule on the second term)

h′′(x)=−4sinx−2xcosx

h′′(π)=−4sin(π)−2(π)cos(π) (subt. x=π)

h′′(π)=6.28

Answer 2

Answer:

Value of $h″(π) \: is \: \pi$.

Step-by-step explanation:

Given

$h(x) = 2xcosx$

Doing derivative with respect to x

h'(x)$= \frac{d}{dx} (2x \cos(x) )$

$= \frac{d}{dx} (2x)cosx + 2x \frac{d}{dx} ( \cos(x) )$

$= 2 \cos(x) - 2xsinx$

Again we are doing derivative with respect to x,

h''(x)$= \frac{d}{dx} (2 \cos(x) - 2x \sin(x))$

$= 2 \frac{d}{dx} ( \cos(x) ) - \frac{d}{dx} (2x) \sin(x) - 2x \frac{d}{dx} ( \sin(x) )$

=$- 2 \sin(x) - 2 \sin(x) - 2x \cos(x)$

Now we want to find h″(π)

We are putting x=(π) in h″(x) then get,

So,h″(π)$= - 2 \sin(\pi) - 2 \sin(\pi) - 2\pi \cos(\pi)$

$- 2 \times 0 - 2 \times 0 - \pi \times ( - 1) = 0 - 0 + \pi = \pi$

So,$h″(π) = \pi$