Question

Differentiate the function
${x}^{cosx} + (cos {x})^{x}$
please, help​

Answer 1

Answer:

$\sf \dfrac{dy}{dx} =x^{cos\:x} \bigg(\dfrac{cos\:x}{x} -sin\:x\:log\:x \bigg)+(cos\:x)^x\bigg(log\:cos\:x-x\:tan\:x \bigg)$

Step-by-step explanation:

Given:

$\sf y= x^{cos\:x}+(cos\:x)^x$

To Find:

dy/dx

Solution:

Let us assume that,

$\sf u=x^{cos\:x}$ and

$\sf v=(cos\:x)^x$

Hence,

$\sf \dfrac{dy}{dx} =\dfrac{du}{dx} +\dfrac{dv}{dx}---(1)$

Finding du/dx

We know that,

$\sf u=x^{cos\:x}$

Taking log on both sides we get,

$\sf log\:u=log\:x^{cos\:x}$

log u = cos x log x

Now differentiate on both sides with respect to x using chain rule,

$\sf\dfrac{1}{u} \:\dfrac{du}{dx} =\dfrac{1}{x}\times cos\:x+log\:x\times -sin\:x$

$\sf \dfrac{du}{dx} =u\bigg(\dfrac{cos\:x}{x} -sin\:x\:log\:x \bigg)$

Substitute the value of u,

$\sf \dfrac{du}{dx} =x^{cos\:x}\bigg(\dfrac{cos\:x}{x} -sin\:x\:log\:x \bigg)---(2)$

Now finding dv/dx

We know,

$\sf v=(cos\:x)^x$

Taking log on both sides,

log v = x log cos x

Differentiate on both sides w.r.t x using chain rule,

$\sf \dfrac{1}{v} \: \dfrac{dv}{dx} =x\times \dfrac{1}{cos\:x} \times -sin\:x+log\:cos\:x\times 1$

$\sf \dfrac{1}{v} \: \dfrac{dv}{dx} =x\times -tan\:x+log\:cos\:x$

$\sf \dfrac{dv}{dx} =v\bigg(log\:cosx-x\:tan\:x \bigg)$

Give back the value of v,

$\sf \dfrac{dv}{dx} =(cos\:x)^x\bigg(log\:cosx-x\:tan\:x \bigg)---(3)$

Substitute 2 and 3 in equation 1,

$\sf \dfrac{dy}{dx} =x^{cos\:x} \bigg(\dfrac{cos\:x}{x} -sin\:x\:log\:x \bigg)+(cos\:x)^x\bigg(log\:cos\:x-x\:tan\:x \bigg)$

Answer 2

$y = x \cos(x) + ( \cos \\ x)$

$now$