Question

Differentiate sinx^x° with respect to x

Answer 1

Answer:

Here

Step-by-step explanation:

Please solve it by yourself

Answer 2

Let y = sin x^x
So,
dy/dx = d(sin x^x)/dx^x * d(x^x)/dx ....by chain rule
So,
dy/dx = cos x^x * d(x^x)/dx

Now let
u = x^x
Taking natural log both sides

ln u = x ln x
Differentiating both sides.

1/u*du/dx = x*1/x + ln x * 1
Therefore,
du/dx = u*(1+ln x) = x^x ( 1 + ln x)
Now, du/dx = d(x^x)/dx = x^x(1+lnx)

Therefore,
d(sin x^x)/dx = x^x(1+ ln x)* cos x^x