Question

Find the derivative of the function
x^x( cotx)^x w.r.t'.x'​

Answer 1

Answer:

$x^{x+1} t^{x-1}.(-cosecx) + cotx^{x}(x^{x})$

Step-by-step explanation:

Let y = $x^{x} .cotx^{x}$

$\frac{dy}{dx} = \frac{d}{dx}(x^{x} .cotx^{x})$

Let

$u = x^{x}\\ v = cotx^{x}$

Since $v = cotx^{x}$ is a compound function.

So, using chain rule, we get

$t = cotx \\v = t^{x}\\\frac{dv}{dx} = \frac{dv}{dt} .\frac{dt}{dx}\\\frac{dv}{dx} = \frac{d}{dt}(t^{x} ) .\frac{d}{dx}(cotx)\\\\\frac{dv}{dx} = x.t^{x-1} .(-cosecx)$

Using product rule, we get

$\frac{d}{dx} (uv) = u\frac{dv}{dx} + v\frac{du}{dx} \\\frac{d}{dx} (uv) = x^{x} \frac{d}{dx}(cotx^{x}) + cotx^{x}\frac{d}{dx}(x^x) \\ \frac{dy}{dx} = x^{x}(x.t^{x-1} .(-cosecx)) + cotx^{x}(x.x^{x-1}) \\ \frac{dy}{dx} = x^{x+1} t^{x-1}.(-cosecx) + cotx^{x}(x^{x-1+1})\\\frac{dy}{dx} = x^{x+1} t^{x-1}.(-cosecx) + cotx^{x}(x^{x})$

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