Find the nth derivative of (sin x)^3

1 answer:

5 0

You'll need to try to find the pattern that applies here. Find the 1st, 2nd, 3rd, 4th and 5th derivatives. Is a pattern evident?

n=1: first derivative of (sin x)^3 is 3(sin x)^2*cos x (which is a product)

n=2: second deriv. of (sin x)^3 is 3(sin x)^2*(-sin x) + (cos x)*(6sin x cos x)

Please find the 3rd, 4th and 5th derivatives. Then try to recognize a pattern in the derivatives governed by the counter "n".

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Can someone help me with this

mash [69]

The center of the circle is (h,k) = (-2,-3)

The radius of the circle is r = 2

The standard form of equation of the circle is

${(x + 2)}^{2} + {(y + 3)}^{2} = 4$

<h3>How to find the center, radius and standrad form of the circle?</h3>

The general form of equation of the circle is

${(x - h)}^{2} + {(y - k)}^{2} = {r}^{2}$

Here, (h,k) means centre of the circle.

r means radius of the circle.

given that coordinate points of centre of circle is (-2,-3).

Hence the (h,k) = (-2,-3)

<h3>How to find the radius of the circle?</h3>

Now to find the radius of the circle

The distance from a circle's centre to its circumference is its radius.

The distance from a circle's centre (-2,-3) to its circumference (0,-3) is its radius.

using the formula, distance between the two points to obtain radius.

$d = \sqrt{(x1 - x2) {}^{2} + {(y1 - y2)}^{2} } \\ r = \sqrt{ {( - 2 - 0)}^{2} + {( - 3 - ( - 3))}^{2} } \\ r = \sqrt{ {( - 2)}^{2} + {( - 3 + 3)}^{2} } \\ r = \sqrt{ {4}^{2} + 0 } \\ r = \sqrt{4} \\ r = 2$