Question

Find the derivative of x5-cosx/sinx​

Answer 1

d/dx {(x⁵ - cosx) / sinx}

= - x⁵ cosecx cotx + 5x⁴ cosecx - cosec²x

Step-by-step explanation:

Let y = (x⁵ - cosx) / sinx

= (x⁵ / sinx) - (cosx / sinx)

= x⁵ cosecx - cotx

Now differentiating both sides w.r. to x, we get

dy / dx = d/dx (x⁵ cosecx - cotx)

= d/dx (x⁵ cosecx) - d/dx (cotx)

= x⁵ d/dx (cosecx) + cosecx d/dx (x⁵) + d/dx (cotx)

= x⁵ (- cosecx cotx) + cosecx (5x⁴) + (- cosec²x)

= - x⁵ cosecx cotx + 5x⁴ cosecx - cosec²x

Rules:

• d/dx (xⁿ) = n xⁿ⁻¹, where n is rational

• d/dx (cosecx) = - cosecx cotx

• d/dx (cotx) = - cosec²x

Related question:

Differentiate Sin inverse of 2 X upon oneplus x square with respect to Cos inverse of 1 - x square upon oneplus x square. - https://brainly.in/question/3911961

Answer 2

$5x^{4}\csc x - x^{5} \cot x \csc x+ \csc^{2}x$

Step-by-step explanation:

We have to find the differentiation of $\frac{x^{5} - \cos x}{\sin x}$ with respect to x.

Now, $\frac{d}{dx}[\frac{x^{5} - \cos x}{\sin x}]$

= $\frac{(\sin x) \frac{d}{dx}(x^{5} - \cos x) - (x^{5} - \cos x)\frac{d}{dx}(\sin x) }{\sin^{2}x }$

{Since we know that the formula of derivative that, $\frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^{2} }$}

= $\frac{(\sin x)(5x^{4} + \sin x) - (x^{5} - \cos x)(\cos x) }{\sin^{2}x }$

{Since the formulas of derivative are, $\frac{d}{dx}(\sin x) = \cos x$, $\frac{d}{dx}(\cos x) = - \sin x$ and $\frac{d}{dx}(x^{n}) = nx^{n - 1}$ }

= $\frac{5x^{4}\sin x + \sin^{2}x - x^{5} \cos x + \cos^{2}x }{\sin^{2}x }$

= $5x^{4}\csc x - x^{5} \cot x \csc x+ \csc^{2}x$ (Answer)