How do you use the chain rule to differentiate $y = {(\frac{cos x}{1 + sin x})}^{5}$ $y=(cosx/(1+sinx))^5$ ?

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How do you use the chain rule to differentiate $y = {(\frac{cos x}{1 + sin x})}^{5}$ $y=(cosx/(1+sinx))^5$ ?

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Answer 1 · 2017-09-20T19:23:00

$- 5 \frac{{cos}^{4} x}{{(1 + sin x)}^{5}}$ $-5(cos^4x)/(1+sinx)^5$

Explanation:

$y = {(\frac{cos x}{1 + sin x})}^{5}$ $y=(cosx/(1+sinx))^5$
$\frac{d y}{d x} = \frac{d}{d x} ({(\frac{cos x}{1 + sin x})}^{5})$ $dy/dx=d/dx((cosx/(1+sinx))^5)$
$= 5 {(\frac{cos x}{1 + sin x})}^{4} \cdot ⎛ ⎜ ⎝ \frac{(1 + sin x) \cdot (\frac{d}{d x} (cos x)) - cos x \frac{d}{d x} (1 + sin x)}{{(1 + sin x)}^{2}} ⎞ ⎟ ⎠$ $=5(cosx/(1+sinx))^4*(((1+sinx)*(d/dx(cosx))-cosxd/dx(1+sinx))/(1+sinx)^2)$

$= 5 {(\frac{cos x}{1 + sin x})}^{4} \cdot (\frac{(1 + sin x) \cdot (- sin x) - cos x (cos x)}{{(1 + sin x)}^{2}})$ $=5(cosx/(1+sinx))^4*(((1+sinx)*(-sinx)-cosx(cosx))/(1+sinx)^2)$

$= 5 {(\frac{cos x}{1 + sin x})}^{4} \cdot (\frac{- sin x - {sin}^{2} x - {cos}^{2} x}{{(1 + sin x)}^{2}})$ $=5(cosx/(1+sinx))^4*((-sinx-sin^2x-cos^2x)/(1+sinx)^2)$

$= 5 {(\frac{cos x}{1 + sin x})}^{4} \cdot (\frac{- sin x - ({sin}^{2} x + {cos}^{2} x)}{{(1 + sin x)}^{2}})$ $=5(cosx/(1+sinx))^4*((-sinx-(sin^2x+cos^2x))/(1+sinx)^2)$

$= 5 {(\frac{cos x}{1 + sin x})}^{4} \cdot (\frac{- sin x - 1}{{(1 + sin x)}^{2}})$ $=5(cosx/(1+sinx))^4*((-sinx-1)/(1+sinx)^2)$

$= 5 {(\frac{cos x}{1 + sin x})}^{4} \cdot \frac{- 1}{1 + sin x}$ $=5(cosx/(1+sinx))^4*(-1)/(1+sinx)$
$= - 5 \frac{{cos}^{4} x}{{(1 + sin x)}^{5}}$ $=-5(cos^4x)/(1+sinx)^5$

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How do you use the chain rule to differentiate $y = {(\frac{cos x}{1 + sin x})}^{5}$ $y=(cosx/(1+sinx))^5$ ?

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Explanation:

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How do you use the chain rule to differentiate $y = {(\frac{cos x}{1 + sin x})}^{5}$ $y=(cosx/(1+sinx))^5$ ?