Question

$\bold \red{Find \: \frac{dy}{dx} \: if \: y= \cos^{5}x}$


❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​

Answer 1

Differentiate separately both sides of the equation (treat y as a function of x)

$\frac{d}{dx}$(y((x)=$\frac{d}{dx}$$(cos^5(x))$

Differentiate the RHS of the equation.

The function cos⁵(x) is the composition f(g(x)) of two functions f(u)=u⁵ and g(x)=cos(x).

Apply the chain rule

${ \bold \red{\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)}}$

${ \bold \red{\color{red}{\left(\frac{d}{dx} \left(\cos^{5}{\left(x \right)}\right)\right)} = \color{red}{\left(\frac{d}{du} \left(u^{5}\right) \frac{d}{dx} \left(\cos{\left(x \right)}\right)\right)}}}$

Apply the power rule

$\bold \red{\frac{d}{du} \left(u^{n}\right) = n u^{n - 1} \: with \: n = 5}$

$\bold{\color{red}{\left(\frac{d}{du} \left(u^{5}\right)\right)} \frac{d}{dx} \left(\cos{\left(x \right)}\right) = \color{red}{\left(5 u^{4}\right)} \frac{d}{dx} \left(\cos{\left(x \right)}\right)}$

Return to the old variable:

${\bold \red{ 5\color{red}{\left(u\right)}^{4} \frac{d}{dx} \left(\cos{\left(x \right)}\right) = 5 \color{red}{\left(\cos{\left(x \right)}\right)}^{4} \frac{d}{dx} \left(\cos{\left(x \right)}\right)}} \\ \bold \red{The \: derivative \: of \: the \: cosine \: is \: \frac{d}{dx} \left(\cos{\left(x \right)}\right) = - \sin{\left(x \right)}}$

${ \bold \red{5 \cos^{4}{\left(x \right)} \color{red}{\left(\frac{d}{dx} \left(\cos{\left(x \right)}\right)\right)} = 5 \cos^{4}{\left(x \right)} \color{red}{\left(- \sin{\left(x \right)}\right)}}}$

Thus,

${\bold \red{\frac{d}{dx} \left(\cos^{5}{\left(x \right)}\right) = - 5 \sin{\left(x \right)} \cos^{4}{\left(x \right)}}}$

Therefore,

$\bold \red{\frac{dy}{dx} = - 5 \sin{\left(x \right)} \cos^{4}{\left(x \right)}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:y = {cos}^{5}x$

can be rewritten as

$\rm :\longmapsto\:y = {(cosx)}^{5}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx} {(cosx)}^{5}$

We know

$\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 5 {(cosx)}^{5 - 1}\dfrac{d}{dx} cosx$

We know

$\boxed{ \tt{ \: \dfrac{d}{dx} cosx \: = \: - \: sinx \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 5 {(cosx)}^{4} \times ( - sinx)$

$\rm \implies\:\boxed{ \tt{ \: \dfrac{d}{dx} {cos}^{5}x \: = \: - \: 5 \: {cos}^{4}x \: sinx \: }}$

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Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$