# Thread: Implicit and log differentiation

1. ## Implicit and log differentiation

Find y' given the following function:
$y=\ln{(5x^2+y^2)}$
So far, I have:
$y'=\frac{10x+2y*y'}{5x^2+y^2}$
Having trouble combining the y's.

Another one I am stuck on is:
$y=(\ln{(x)})^{cos(15x)}$
I am not positive if it is okay to bring the cos(15x) into the front of the ln in this case?
Thanks for any help!

2. Originally Posted by xxlvh
Find y' given the following function:
$y=\ln{(5x^2+y^2)}$
So far, I have:
$y'=\frac{10x+2y*y'}{5x^2+y^2}$
Having trouble combining the y's.

Multiply by the denominator:

y'(5x^2+y^2) = 10x + 2yy' ==> y'(y^2 - 2y + 5x^2) = 10x and etc.

Another one I am stuck on is:
$y=(\ln{(x)})^{cos(15x)}$
I am not positive if it is okay to bring the cos(15x) into the front of the ln in this case?
Thanks for any help!

Of course not: you can only "bring down" the power of the argument, NOT of the whole logarithm, i.e.:
log(x^n) = n*log x, but (log x)^n =/= n*log x

You better write y = e^[(cos15x)ln(x)] and derive now using the chain rule.

Tonio

3. How exactly did you rearrange it to the form y = e^[(cos15x)ln(x)]?
And I was expected to use logarithmic differentiation to solve this particular one, without being able to bring that cos(15x) into the front though I'm not sure how to go about it.

4. Originally Posted by xxlvh
How exactly did you rearrange it to the form y = e^[(cos15x)ln(x)]?
And I was expected to use logarithmic differentiation to solve this particular one, without being able to bring that cos(15x) into the front though I'm not sure how to go about it.

Well, for any positive a and any real x, the identity a^x = e^(x*ln(a)) is pretty easy to check.
In our case it perhaps makes it easier to differentiate since the derivative of the exponential function is very easy......and still you need a logarithmic differentiation.

You can try to differentiate directly though, but I'm afraid it's gonna get nasty. I, for one, would never, ever do it directly.

Tonio

5. I must still be making an error.
If I use
$y=e^{cos(15x)\ln{x}}$

then

$y' = (cos(15x)ln(x))e^{(cos(15x)ln(x))-1}(\frac{cos(15x)}{x}-15sin(15x)ln(x))$ ???

When graphed, it does not match the derivative of the original function $y = (ln(x))^{cos(15x)}$

6. Originally Posted by xxlvh
I must still be making an error.
If I use
$y=e^{cos(15x)\ln{x}}$

then

$y' = (cos(15x)ln(x))e^{(cos(15x)ln(x))-1}(\frac{cos(15x)}{x}-15sin(15x)ln(x))$ ???

When graphed, it does not match the derivative of the original function $y = (ln(x))^{cos(15x)}$
Let's do it as follows:

[e^(cos15x lnx)]' = [cos15x lnx]'*e^(cos15x lnx) =

[-15sin15x lnx + cos15x/x]*ln(x)^(cox15x) , and we're done.

I can't understand from where did you bring that -1 in the power of e. Apparently you confused something here and tried to derivate something as a polynomial : (x^n)' = n x^(n-1)...or stuff.

Tonio

7. Whoops! Yes that was my mistake.
However, your solution doesn't yield the proper graph for the derivative either.

8. Originally Posted by xxlvh
Whoops! Yes that was my mistake.
However, your solution doesn't yield the proper graph for the derivative either.

The derivative of e^f(x) is f '(x)*e^f(x) , when f(x) is a derivable function.
This is what I did above, and I don't know what you mean by "your solution doesn't yield the proper graph...? What proper graph of what??

Tonio

9. I am graphing the derivative of the original function and the solution you just posted for the derivative. They don't match up.

On top of that, I'm entering that in for the derivative (it's an online assignment) and it's still getting marked as incorrect..I've checked it several times to make sure I'm not making mistakes typing it in.

10. Originally Posted by xxlvh
I am graphing the derivative of the original function and the solution you just posted for the derivative. They don't match up.

On top of that, I'm entering that in for the derivative (it's an online assignment) and it's still getting marked as incorrect..I've checked it several times to make sure I'm not making mistakes typing it in.
I just found the mistake: the function is e^[cos(15x)*ln(lnx)], since the
basis is lnx, not x!
Well, do as before with the new f(x) = cos(15x)ln(lnx).

Tonio