Find the derivative of (x^2+1)^x.

Question

zepdrix · Answer

Ooo this one is a stinker!!
We have our variable term in the base AND in the exponent.
So neither the power rule, nor the rule for exponents will work.

We should use logs to get through this.

anonymous · Answer

take log and then diff.

anonymous · Answer

Can you show the work?

zepdrix · Answer

$$\Large\bf\sf y=(x^2+1)^x\qquad\to\qquad \ln y=\ln(x^2+1)^x$$Remember this rule for logs?$$\Large\bf\sf \color{orangered}{\log(a^b)\quad=\quad b\cdot \log(a)}$$

anonymous · Answer

So ln y=x*ln(x^2+1)?

zepdrix · Answer

good good.

zepdrix · Answer

Now differentiate (with respect to x), applying the product rule on the right side.

anonymous · Answer

But on the left side, it will be (1/y)(dy/dx), right?

zepdrix · Answer

Yes, good :)

anonymous · Answer

Okay, so I got (1/y)(dy/dx)=2x^2/(x^2+1)+ln(x^2+1). But how do I simplify this by multiply y to the right side? What will it look like?

zepdrix · Answer

Ya multiply y to the other side, then we want to substitute our original equation into that y.$$\Large\bf\sf y=(x^2+1)^x$$

anonymous · Answer

So how does it look like?

zepdrix · Answer

We want our derivative in terms of `x` alone.

zepdrix · Answer

$$\Large\bf\sf y'=(x^2+1)^x\left[\ln(x^2+1)+\frac{2x^2}{x^2+1}\right]$$

zepdrix · Answer

That is a good place to stop. Good looking final answer.

You COULD find a common denominator and then it simplifies down a little bit further.
But it's not really necessary.

anonymous · Answer

I got it. Exactly the answer I was looking for. Thanks a lot.

zepdrix · Answer

Cool \c:/ good job