Question

differentiate
((acosx + bsinx + c) / sinx)+ x^n cos x​

Answer 1

Topic :-

Differentiation

To Differentiate :-

$y=\dfrac{a\cos x+b\sin x+c}{\sin x}+x^n\cdot \cos x$

Solution :-

$y=\dfrac{a\cos x+b\sin x+c}{\sin x}+x^n\cdot \cos x$

$y=\dfrac{a\cos x}{\sin x}+\dfrac{b\sin x}{\sin x}+\dfrac{c}{\sin x}+x^n\cdot \cos x$

$y=a\cot x+b+c\csc x+x^n\cdot \cos x$

Differentiating both sides,

$\dfrac{dy}{dx}=\dfrac{d(a\cot x)}{dx}+\dfrac{d(b)}{dx}+\dfrac{d(c\csc x)}{dx}+\dfrac{d(x^n\cdot \cos x)}{dx}$

$\dfrac{dy}{dx}=a\dfrac{d(\cot x)}{dx}+\dfrac{d(b)}{dx}+c\dfrac{d(\csc x)}{dx}+\dfrac{d(x^n\cdot \cos x)}{dx}$

$\left( \because \dfrac{d(k.f(x))}{dx}=k\dfrac{d(f(x))}{dx},where\:k\:is\:a\:constant.\right)$

$\dfrac{dy}{dx}=-a\csc^2x+\dfrac{d(b)}{dx}+c\dfrac{d(\csc x)}{dx}+\dfrac{d(x^n\cdot \cos x)}{dx}$

$\left( \because \dfrac{d(\cot x)}{dx}=-\csc^2x\right)$

$\dfrac{dy}{dx}=-a\csc^2x+0+c\dfrac{d(\csc x)}{dx}+\dfrac{d(x^n\cdot \cos x)}{dx}$

$\left( \because \dfrac{d(b)}{dx}=0,where\:b\:is\:a\:constant.\right)$

$\dfrac{dy}{dx}=-a\csc^2x-c\csc x.\cot x+\dfrac{d(x^n\cdot \cos x)}{dx}$

$\left( \because \dfrac{d(\csc x)}{dx}=-\csc x.\cot x\right)$

$\dfrac{dy}{dx}=-a\csc^2x-c\csc x.\cot x+\cos x.\dfrac{d(x^n)}{dx}+x^n.\dfrac{d(\cos x)}{dx}$

$(\because (fg)'=gf'+fg')$

$\dfrac{dy}{dx}=-a\csc^2x-c\csc x.\cot x+nx^{n-1}.\cos x+x^n.\dfrac{d(\cos x)}{dx}$

$\left( \because \dfrac{d(x^n)}{dx}=nx^{n-1}\right)$

$\dfrac{dy}{dx}=-a\csc^2x-c\csc x.\cot x+nx^{n-1}.\cos x-x^n.\sin x$

$\left( \because \dfrac{d(\cos x)}{dx}=-\sin x\right)$

Answer :-

$\dfrac{dy}{dx}=-a\csc^2x-c\csc x.\cot x+nx^{n-1}.\cos x-x^n.\sin x$