Question

Find the derivative of x^x + (sinx)^x

.​

Answer 1

Answer:

$\tt \dfrac{dy}{dx}=x^x(log\:x+1)+(sin\:x)^x(log\:sin\:x+x\:cot\:x)$

Step-by-step explanation:

Given:

$\tt y=x^x+(sin\:x)^x$

To Find:

$\tt \dfrac{d}{dx} (y)$

Solution:

By given,

$\tt y=x^x+(sin\:x)^x$

Now let,

$\tt u=x^x$  and

$\tt v=(sin\:x)^x$

Hence,

$\tt \dfrac{d}{dx} (y)=\dfrac{d}{dx} (u)+\dfrac{d}{dx}(v)---(1)$

Consider,

$\tt u=x^x$

Taking log on both sides,

$\tt log\:u=log\: x^x$

We know that,

$\tt log\:a^m=m\:log\:a$

Therefore,

$\tt log\:u=x\:log\:x$

Now differentiate on both sides with respect to x.

Using product rule,

$\tt (uv)'=u'v+v'u$

Also,

$\tt \dfrac{d}{dx} (x)=1$

$\tt \dfrac{d}{dx} (log\:x)=\dfrac{1}{x}$

Hence,

$\tt \dfrac{1}{u} \:\dfrac{du}{dx}=1\times log\:x+x\times \dfrac{1}{x}$

$\tt \dfrac{1}{u} \:\dfrac{du}{dx}= log\:x+1$

$\tt \dfrac{du}{dx} =(log\:x+1)\:u---(2)$

Now consider,

$\tt v=(sin\:x)^x$

Take log on both sides and simplify,

$\tt log\:v=x\: log\:sin\:x$

Differentiate with respect to x using product rule and chain rule,

We know,

$\tt \dfrac{d}{dx} (sin\:x)=cos\:x$

Hence,

$\tt \dfrac{1}{v}\:\dfrac{dv}{dx}=1\times log(sin\:x)+x\times \dfrac{1}{sin\:x}\times cos\:x$

$\tt \dfrac{1}{v}\:\dfrac{dv}{dx}= log(sin\:x)+x\times cot\:x$

$\tt \dfrac{dv}{dx}= (log(sin\:x)+x\times cot\:x)\: v---(3)$

Substitute 2 and 3 in 1,

$\tt \dfrac{dy}{dx}=u(log\:x+1)+v(log(sin\:x)+x\:cot\:x)$

Give back the value of u and v,

$\tt \dfrac{dy}{dx}=x^x(log\:x+1)+(sin\:x)^x(log\:sin\:x+x\:cot\:x)$

Answer 2

Correct question= What is the derivative of x^ (sinx)?

Solution=⬇️

y=xsinx

Method 1:

Taking log on both sides

logy=sinxlogx

Differentiating both sides using product rule (for y = uv where both u and v are functions of x)

1ydydx=sinxx+cosxlogx

Taking y to RHS we get

dydx=xsinx[sinxx+cosxlogx]

Method 2:

This is similar to the product tule of differentiation. Basically in product rule what we do is, we treat u constant and differentiate v and then take v constant and differentiate u and add them together. Similarly if we take this function, it is of the form u^v. So when we take u as constant, it becomes a differentiation of the form a^x (which will give a^x lna) (also dont forget to differentiate sinx) and then when we take v as a constant, it will be of the form x^n (which will give nx^(n-1) ).

Try this method on your own as practice. In case you dont get, I will write the solution.

Hope it helps.