use logarithmic differentiation to find the derivative of the
function. y=(x^3+2)^2(x^5+4)^4

Question

use logarithmic differentiation to find the derivative of the 
function. y=(x^3+2)^2(x^5+4)^4

terenzreignz · Answer

$$y=\left( x^{3}+2 \right)^{2}\left( x^{5}+4 \right)^{4}$$
Is that right?

terenzreignz · Answer

While you don't technically have to use logarithmic differentiation here, evaluating the powers of these binomials and then multiplying said powers and only then differentiating normally... might be a pain.. so let's get started :)

$$\ln y=\ln \left[ \left( x^{3}+2 \right)^{2}\left( x^{5}+4 \right)^{4} \right]$$

anonymous · Answer

I agree with your comment. Y use log?

anonymous · Answer

I was thinking of using implicit as well

terenzreignz · Answer

Following the properties of the logarithm:
$$\ln y = \ln \left( x^{3}+2 \right)^{2}+\ln \left( x^{5}+4 \right)^{4}$$
And yet more properties of logarithms...
$$\ln y = 2\ln \left( x^{3}+2 \right)+4\ln \left( x^{5}+4 \right)$$

anonymous · Answer

So far so good. I get it yay

terenzreignz · Answer

Use log because... it has this awesome ability to turn exponents into factors :D Also to turn products into sums... something like that

terenzreignz · Answer

Can you work it out from there?

anonymous · Answer

Hmm not sure about what u just said. But sound awesome. Lol

anonymous · Answer

Please help some more

terenzreignz · Answer

Well look at it: the 2 and the 4 which were once exponents, now they are just constants multiplied to logs... also, those two binomials were multiplied to each other, but now they are the sum of logs :)

If nothing else, use logs because the instructions say so :P

terenzreignz · Answer

Maybe later, I'll show you a more practical use of logs, but for now...
See that last expression there?  Differentiate it implicitly....

anonymous · Answer

What last expression thw one u sent earlier

anonymous · Answer

If u can provide answer. I will get my own and match it to urs. Btw thanks so much
U r awesome

terenzreignz · Answer

So you should get
$$\frac{1}{y}\frac{dy}{dx}=\left( \frac{2}{x^{3}+2} \right)\left( 3x^{2} \right)+\left( \frac{4}{x^{5}+4} \right)\left( 5x^{4} \right)$$

terenzreignz · Answer

Now it's easier to see the derivative... don't you agree?

terenzreignz · Answer

Anyway, to get the derivative, just take that expression there on the right, and multiply it to y, which is given as a function of x, as in my first comment.  And you're done!

Also, there are such functions, like
$$y = x^{x}$$
for which I don't know how to get the derivative except via logarithmic differentiation.

anonymous · Answer

Yes i got this as well. Yay but i c the reason behind log derivatives thanks again

terenzreignz · Answer

No sweat