Question

How do you find the derivative of `log_5(arctanx^x)`?

Answer 1

`("d")/("d"x) log_5(arctan(x^x)) = ((1+ln(x))x^x)/(ln(5)(x^{2x}+1)arctan(x^x))`.

Explanation:

`y=log_5(arctan(x^x))`.

The right hand side of this expression is fairly horrible. It would be better to rearrange this and use implicit differentiation than attempt to differentiate directly.

`5^y = arctan(x^x)`.

A common trick to help differentiate functions involving exponents is to write the base as `e` and modify the power with a logarithm to give the desired exponent. This then makes differentiating easier. Notice that `x^x=e^{xln(x)}` and `5^y=e^{yln(5)}`. Then,

`e^{yln(5)}=arctan(e^{xln(x)})`.

Differentiate implicitly. Note that `"d"/("d"x) arctan(x) = 1/(x^2+1)`.

`ln(5)e^{yln(5)}("d"y)/("d"x)=1/(e^{2xln(x)}+1) "d"/("d"x) (e^{xln(x)})`.

As `"d"/("d"x) e^{xln(x)} = (1+ln(x))e^{xln(x)}`, we see that,

`("d"y)/("d"x) ln(5) e^{yln(5)} =1/(e^{2xln(x)}+1) (1+ln(x))e^{xln(x)}`.

We can now rewrite the exponents as what they were originally and rearrange for `("d"y)/("d"x)`.

`("d"y)/("d"x) = ((1+ln(x))x^x)/((x^{2x}+1)ln(5)5^y)`.

We know from the definition that `5^y=arctan(x^x)`.

We conclude,

`("d"y)/("d"x) = ((1+ln(x))x^x)/(ln(5)(x^{2x}+1)arctan(x^x))`.