Question

Solve the derivative by quotient rule :-

$y \: = \: \frac{ {x}^{5} - \: cosx}{sinx}$
​

Answer 1

Quotient rule implies,

$\boxed{\begin{minipage}{9.83 cm}\large \begin{aligned}\textsf{If}\ \ \ &\ \ \ y=\dfrac{f(x)}{g(x)}\ , \\ \\ \textsf{then}&\ \ \ y'=\dfrac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}\end{aligned}\end{minipage}}$

Given that,

$y=\dfrac{x^5-\cos x}{\sin x}$

From this, we get,

$\rightarrow\ f(x)=x^5-\cos x\\ \\ \rightarrow\ g(x)=\sin x$

Now we find  f'(x)  and  g'(x).  

$\rightarrow\ f'(x)=5x^4+\sin x\\ \\ \rightarrow\ g'(x)=\cos x$

So,

$\begin{aligned}y'&=\ \ \frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}\\ \\ y'&=\ \ \frac{\sin x(5x^4+\sin x)-(x^5-\cos x)\cos x}{\sin^2x}\\ \\ y'&=\ \ \frac{5x^4\sin x+\sin^2x-x^5\cos x+\cos^2x}{\sin^2x}\\ \\ \\ \Large y'&=\ \ \Large \frac{-x^5\cos x+5x^4\sin x+1}{\sin^2x}\end{aligned}$

Hence Derived!

Answer 2

$\sf{Let\:f(x)={\dfrac{{x}^{5} - \: cosx}{sinx}}}$

$\sf {Let\:u= x^{5} - cosx\:and\:v = sinx}$

$\sf {f(x) = ({\dfrac{u}{v}})}$

•°• $\sf {f^{'}(x) = {\dfrac{u}{v}}^{'}}$

$\sf{Using\:Quotient\:Rule,}$

$\sf {f^{'}(x) = ({\dfrac{u^{'}v - v^{'}u}{v^{2}}})}$

$\sf{\underline{Finding\:u^{'}\:and\:v^{'}}}$

$\sf{u = x^{5} - cosx}$

$\sf{u^{'} = 5 . x^{5 - 1} - ( - sinx )}$

$\sf{(Derivative\:of\:x^{n}\:is\:nx^{n-1}\:and\: Derivative\: of \:cosx = - sinx)}$

$\sf {v = sinx}$

$\sf{v^{'} = cos x}$ $\sf{(Derivative\:of\:sinx = cosx)}$

$\sf{Now,}$

$\sf {f^{'}(x) = ({\dfrac{u}{v}})^{'}}$

$\sf{= {\dfrac{u^{'}v - v^{'}u}{v^{2}}}}$

$\sf {= \dfrac {(5x^{4} + sinx) sinx - (cosx)(x^{5} - cosx)}{sin^{2}x}}$

$\sf {= \dfrac {5x^{4} sinx + sin^{2}x - cosx . x^{5} + cos^{2}x}{sin^{2}</p><p>x}}$

$\sf {= \dfrac {- x^{5} cosx + 5x^{4}sinx + sin^{2}x + cos^{2}x}{sinx^{2}}}$

$\sf {= \dfrac {- x^{5} cosx + 5x^{4}sinx + 1}{sinx^{2}}}$ $\sf {(sin^{2}x + cos^{2}x = 1)}$

$\sf {(sin^{2}x + cos^{2}x = 1)}$

$\sf{Thus, f^{'}(x) = \dfrac {- x^{5} cosx + 5x^{4}sinx + 1}{sinx^{2}}}$