1. ## Derivatives and Anti-Derivatives

Hey everyone. I was wondering if I could get help with these three problems.

Be sure to click the show steps button.

3. Originally Posted by FullBox
Hey everyone. I was wondering if I could get help with these three problems.

For the first one, the answer wouldn't be in an integral.

4. Originally Posted by dwsmith
For the first one, the answer wouldn't be in an integral.
Yeah that was a total typo on my part. You can even see in the formula I typed there was no integral, I don't know why I put it there. My tired-ness is catching up with me. But apart from that integral typo, it seems to be correct according to Plato's link.

And Plato, I typed in the second problem into that website and it said:

This is what I typed into the search box:

derivative of (x^3+1) ^tan x

5. Originally Posted by FullBox
Yeah that was a total typo on my part. You can even see in the formula I typed there was no integral, I don't know why I put it there. My tired-ness is catching up with me. But apart from that integral typo, it seems to be correct according to Plato's link.

And Plato, I typed in the second problem into that website and it said:

This is what I typed into the search box:

derivative of (x^3+1) ^tan x
You can just input the equation directly and wolfram will pull up the integral, limit, derivative, etc of the function without you having to say a word!

Also, you could do this,

Let $\displaystyle y = (x^3+1)^{tanx}$

$\displaystyle lny = tan(x) ln (x^3+1)$

Now we can implicitly differentiate the left side and apply the product rule to the right side. Should clear everything up.

6. Originally Posted by FullBox
And Plato, I typed in the second problem into that website and it said:
Well you are not suppose to need steps for derivatives.

7. $\displaystyle \frac{db^{u(x)}}{dx}= ln(b)b^{u(x)}u'(x)$ is true only for b a constant.
(Note the $\displaystyle b^{u(x)}$, not u(x). What you wrote:
"$\displaystyle ln(b)u(x)u'(x)$ is not true even for b a constant.)

To differentiate $\displaystyle y= f(x)^{g(x)}$ use the fact that $\displaystyle ln(y)= g(x)ln(f(x)$ as Allencuz suggested.

There are two common errors made in differentiating something like $\displaystyle f(x)^{g(x)}$:
1) Treat the base as a constant: $\displaystyle ln(f(x))f(x)^{g(x)}g'(x)$.

2) Treat the exponent as a constant: $\displaystyle g(x)f(x)^{g(x)- 1}f'(x)$.

The interesting thing is that the correct derivative is the sum of those errors!

8. Originally Posted by AllanCuz
You can just input the equation directly and wolfram will pull up the integral, limit, derivative, etc of the function without you having to say a word!

Also, you could do this,

Let $\displaystyle y = (x^3+1)^{tanx}$

$\displaystyle lny = tan(x) ln (x^3+1)$

Now we can implicitly differentiate the left side and apply the product rule to the right side. Should clear everything up.
I don't understand what you did when you played ln on both sides [what property did you use?]. I did, however do as you said and applied the product rule to the right side.

Originally Posted by HallsofIvy
$\displaystyle \frac{db^{u(x)}}{dx}= ln(b)b^{u(x)}u'(x)$ is true only for b a constant.
(Note the $\displaystyle b^{u(x)}$, not u(x). What you wrote:
"$\displaystyle ln(b)u(x)u'(x)$ is not true even for b a constant.)

To differentiate $\displaystyle y= f(x)^{g(x)}$ use the fact that $\displaystyle ln(y)= g(x)ln(f(x)$ as Allencuz suggested.

There are two common errors made in differentiating something like $\displaystyle f(x)^{g(x)}$:
1) Treat the base as a constant: $\displaystyle ln(f(x))f(x)^{g(x)}g'(x)$.

2) Treat the exponent as a constant: $\displaystyle g(x)f(x)^{g(x)- 1}f'(x)$.

The interesting thing is that the correct derivative is the sum of those errors!
So what you are saying is that x^(3)+1 cannot be treated like, say, pi would for example. So if I had something like pi instead of x^(3)+1, the rule I used earlier would apply?