# f(x)=X^2*log x What is the derivative of this function?

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You need to use the product rule to find the derivative of the function such that:

`f'(x) = (x^2)'*log x + x^2*(log x)'`

You should convert the base of logarithm such that:

`log x = (ln x)/(ln 10) =gt (log x)' = (1/x)/(ln 10)`

`f'(x) = 2x*log x + (x^2)/(x ln 10)`

Reducing by x the second term yields:

`f'(x) = 2x*log x + x/(ln 10)`

**Hence, evaluating the derivative of the given function yields `f'(x) = 2x*log x + x/(ln 10).` **

**Sources:**

First, you will need to know how to get the derivative of the products.

When F(x) = f(x)g(x),

F'(x) = f'(x)g(x) + f(x)g'(x)

The function given in the question is f(x) = x^2*logx

d(x^2)/dx = 2x

d(logx)/dx = lnx/ln10

Therefore, in this case where the function is x^2*logx

f'(x) = d(x^2)/dx*logx + (x^2)*d(logx)/dx

= 2x*logx + (x^2)*lnx/ln10

f(x) f'(x)

a log x 1/(x ln a)

so in this case it will be:

1/ (x ln x^2)

hope this will help