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I can't do any of these problems really. Going through them all I felt really good, I thought they were easy and that I had finally got a grasp on this stuff. I think went to check the answers and I didn't get a single one correct. Anyways I will just pick out a random one, I can't get any of these on my own.

I am sure this one is pretty easy for most people.

differentiate $\sin x \ln(5x)$ I get $\cos x(\ln5x)+\sin x/5x$ which is an incorrect answer. Not sure why.

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2 Answers 2

Apply the chain rule on the second term.

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What is the second term? –  user138246 Oct 13 '11 at 22:20
    
$\frac{d}{dx}[\ln ax]=\frac{1}{ax}.\frac{d}{dx}[ax]=?$ Or, use $\ln(5x)=\ln5+\ln x $ –  Tapu Oct 13 '11 at 22:51
    
I am confused. Why do I not just used ln5x? –  user138246 Oct 14 '11 at 0:50
    
@Jordan: You are trying to do the derivative of $\ln(5x)$ in the very last step of the derivative. Either you have to rewrite it as $(\ln(5x))' = (\ln(5) + \ln(x))'$ using the properties of the logarithm and then take the derivative, or you have to use the Chain Rule, because $\ln(5x)$ is the composition of $f(u)=\ln(u)$ and $g(x) = 5x$. –  Arturo Magidin Oct 14 '11 at 3:37
    
I still don't understand that. I mean if I have 3x I am not going to say that is a composition of 3 and x, I am just going to say the derivative is 3. If I have sin3x I am not going to say it is a composition of sin and 3x, it is sin3x. –  user138246 Oct 14 '11 at 12:26

You seem to be under the impression that the derivative of $$\log({\rm any\ function\ of\ }x) $$ is $$1/({\rm that\ function\ of\ }x)$$ but in fact the chain rule says that it is $$(1/{\rm that\ function\ of\ }x)\times({\rm derivative\ of\ that\ function\ of\ }x)$$

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I am confused because this is not what the forumla in my book is. –  user138246 Oct 14 '11 at 0:49
    
What's the formula in your book? Would it be better if I wrote $(\log(f(x)))'\ne1/(f(x))$, $(\log(f(x)))'=(1/(f(x)))(f'(x))$? –  Gerry Myerson Oct 14 '11 at 1:25
    
My book has the derivative of logax as 1/xlna –  user138246 Oct 14 '11 at 1:27
    
OK, then I think what you are writing here as logax is the logarithm of $x$ to the base $a$, which I'd write as $\log_ax$. It's true that the derivative of $\log_ax$ is $1/(x\log a)$. –  Gerry Myerson Oct 14 '11 at 3:21

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