Question

find the nth derivative of sin^2x cos^3x
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Answer 1

Answer:

using formula are sin^2x=(1-cos2x) /2

cos^3x=(cos3x+3cosx) /4

cosAcosB=1/2(cos(A+B) +cos(A-B))

Answer 2

The require answer is  $\frac{1}{8}$ cos( x+nπ/2) - 3^n/16 cos(3x +nπ/2) -5^n/16 cos(5x +nπ/2).

derivatives: the instantaneous rate of change of a function, and the instantaneous velocity of an object at x=a all required us to compute the following limit.

             $\lim_{n \to \a}$a  $\frac{f(x)-f(a)}{x-a}$.

As per the question we need to find the nth derivative of  sin^2x cos^3x.

nth derivatives of trigonometric functions having powers in them are tough to deal with, which is why we will employ these two power reducing identities,

  $sin^{2}x$ = $\frac{1-cos2x}{2}$

 cos^3(x) =$\frac{3cosx +cos3x}{4}$

to rewrite the given expression as,

$\frac{cosx}{8} -\frac{cos3x}{16} -\frac{cos5x}{16}$

Note that I have used the formula,

cosx cosy=$\frac{1}{2}${cos(x-y) -cos(x+y)}

to convert products of the cosines to sum of cosines.

Now calculating the nth derivative is pretty easy. Use the fact that the nth derivative of cos(ax) is given by,

d/dx^n(cosax) =a^n cos(ax +nπ/2)

and evaluate the nth derivative as,

$\frac{1}{8}$ cos( x+nπ/2) - 3^n/16 cos(3x +nπ/2) -5^n/16 cos(5x +nπ/2)

Hence the require answer is  $\frac{1}{8}$ cos( x+nπ/2) - 3^n/16 cos(3x +nπ/2) -5^n/16 cos(5x +nπ/2).

To learn more about derivative follow the given link

https://brainly.com/question/25324584?

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