# Derivatives and Anti-Derivatives

• Feb 5th 2011, 01:45 PM
FullBox
Derivatives and Anti-Derivatives
• Feb 5th 2011, 02:28 PM
Plato
You can use this website to check your answers.
Be sure to click the show steps button.
• Feb 5th 2011, 02:31 PM
dwsmith
For the first one, the answer wouldn't be in an integral.
• Feb 5th 2011, 02:36 PM
FullBox
Quote:

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:

http://www3.wolframalpha.com/Calcula...=29&w=227&h=20

This is what I typed into the search box:

derivative of (x^3+1) ^tan x
• Feb 5th 2011, 02:54 PM
AllanCuz
Quote:

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:

http://www3.wolframalpha.com/Calcula...=29&w=227&h=20

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

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

Now we can implicitly differentiate the left side and apply the product rule to the right side. Should clear everything up.
• Feb 5th 2011, 02:59 PM
Plato
Quote:

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.
• Feb 6th 2011, 03:57 AM
HallsofIvy
$\frac{db^{u(x)}}{dx}= ln(b)b^{u(x)}u'(x)$ is true only for b a constant.
(Note the $b^{u(x)}$, not u(x). What you wrote:
" $ln(b)u(x)u'(x)$ is not true even for b a constant.)

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

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

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

The interesting thing is that the correct derivative is the sum of those errors!
• Feb 6th 2011, 06:21 AM
FullBox
Quote:

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

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

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.

http://www.mathhelpforum.com/math-he...s-capture5.jpg

Attachment 20697

Quote:

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

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

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

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

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?