Question

Que. 2] Differentiate with respect to if y = x/ sin^x​

Answer 1

Question :

differentiation with respect to x

if $y = \frac{x}{sin{}^{x} x }$

Formula :

1) Quotient Rule

$\frac{d}{dx} (\frac{u}{v} ) = \frac{v \frac{du}{dx} - u \frac{dv}{dx} }{v {}^{2} }$

2) Product rule

$\frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

Solution :

let u = x and v = $sin{}^{x}x$

$v = \sin {}^{x} (x)$

take log on both sides

$log(v) = log( \sin {}^{x} (x) )$

we know that ;

$loga {}^{n} = n log(a)$

$\implies \: log(v) = x log( \sin(x) )$

now by product rule , differentiate it with respect to x

$\implies \frac{1}{v} \times \frac{dv}{dx} = x \times \frac{1}{sinx} \times cosx \: + log( \sin(x) ) \times 1$

$\implies \: \frac{dv}{dx} = v( x \cot(x) + log( \sin(x) )$

$\implies \: \frac{dv}{dx} = \sin {}^{x} (x) (x \cot(x) + log( \sin(x) ) \: \: ..(1)$

$u = x$

$\implies \frac{du}{dx} = 1$

Now

$y = \frac{x}{ \sin(x) }$

by Quotient Rule , difffentiate it with respect to x

$\frac{dy}{dx} = \frac{ \sin {}^{x} (x) \times \frac{dx}{dx} - x \times \frac{d (\sin {}^{x} (x) )}{dx} }{( \sin {}^{x} (x)) {}^{2} }$

put the value of $\frac{d( \sin {}^{x} (x)) }{dx} = \frac{dv}{dx} \: \: \: from \: ...(1)$

${\purple{\boxed{\large{\bold{</p><p> \implies \: \frac{dy}{dx} = \frac{ \sin {}^{x} (x) - x \times \sin {}^{x} (x)(x \cot(x) + log( \sin(x) ) )}{( \sin {}^{x} (x)) {}^{2} } }}}}}$

it is the required solution!