
3 Attachment(s)
Derivatives and AntiDerivatives


Quote:
Originally Posted by
FullBox
For the first one, the answer wouldn't be in an integral.

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 tiredness 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

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 tiredness 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 $\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.

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.

$\displaystyle \frac{db^{u(x)}}{dx}= ln(b)b^{u(x)}u'(x)$ 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!

1 Attachment(s)
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 $\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.
http://www.mathhelpforum.com/mathhe...scapture5.jpg
Attachment 20697
Quote:
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?