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Can someone help me differentiate the function $$\frac{\ln(x)+\sin(x)}{x^2}$$ With respect to $x$? I tryed applying the chain rule but I keep geting lost.

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5  
Where are you stuck? –  Pragabhava May 21 '13 at 18:49
2  
I don't think you need the chain rule here. It seems like we need to apply the quotient rule. –  Jared May 21 '13 at 18:50
    
Do it in pieces to observe that the chain rule is really not necessary. You have a function $f(x)/g(x)$. To differentiate it, use the quotient rule to get $(f'g-gf')/g^2$. Now you just need to figure out what $f'$ and $g'$ are. $g$ is easy. $f$ is the sum of functions. But this is easy too, just differentiate each one –  Dhruv Ranganathan May 21 '13 at 18:53
    
It is better to post with an attempt, or plan, of your own. However, you may simply write the function as $x^{-2}(\ln(x) + \sin(x))$ and use the product rule instead. Though, keep in mind, that you must make the assumption $x > 0.$ –  ThisIsNotAnId May 21 '13 at 19:26

3 Answers 3

Hint:$(\frac{u}{v}) '=\frac{u 'v-uv '}{v^2}$ $$u=\ln(x)+\sin(x) \to u ' =\frac1x+cos(x)$$

$$v=x^2\to v'=2x$$

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You can use the quotient rule to solve this. Given $f(x)=\frac{u}{v}$, $f'(x)=\frac{u'v-uv'}{v^2}$.

So we have:

$$f'(x)=\frac{(\frac{1}{x}+\cos(x))(x^2)-(\ln(x)+\sin(x))(2x)}{x^4}$$

Which simplifies to:

$$f'(x)=\frac{x+x^2\cos(x)-2x\ln(x)-2x\sin(x)}{x^4}$$

And then:

$$f'(x)=\frac{1+x\cos(x)-2\ln(x)-2\sin(x)}{x^3}$$

Alternatively, you can use the product rule ($f(x)=uv \implies f'(x)=uv'+u'v$) for this on the following function. The product rule is often more intuitive than the quotient rule, so this might work better for you.:

$$f(x)=(\ln(x)+\sin(x))(x^{-2})$$

Which gives:

$$f'(x)=(\ln(x)+\sin(x))(-2)(x^{-3})+x^{-2}(\frac{1}{x}+\cos(x))$$

This simplifies to:

$$f'(x)=(\ln(x)+\sin(x))(-2)(x^{-3})+x^{-3}(1+x\cos(x))$$

And then:

$$f'(x)=\frac{-2\ln(x)-2\sin(x)+1+x\cos(x)}{x^{3}}$$

We can see that the solutions from the applications of the quotient rule and the product rule are equal, so that's a great way to check the answer as well!

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Why was this downvoted? –  Pedro Tamaroff May 21 '13 at 19:14
    
@PeterTamaroff I was a bit confused about that too -- I initially answered with just the product rule and then added the quotient rule after. I may have been downvoted before that edit because the quotient rule is a more conventional way to solve the given problem. –  Andy Bromberg May 21 '13 at 19:18

First of all, there's no need to use the chain rule. The chain rule is for when you have a function of a function, for example $\operatorname{e}^{\sin x}$ or $(\sin x - \cos x)^2$. In this case, the quotient rule will suffice. Notice

Recall that the quotient rule says that

$$\left( \frac{\operatorname{f}}{\operatorname{g}} \right)' = \frac{\operatorname{f}'\!\operatorname{g}-\operatorname{f}\!\operatorname{g}'}{\operatorname{g}^2}$$

In your question $\operatorname{f}(x) = \ln x + \sin x$ and $\operatorname{g}(x) = x^2$. Recall that $(\ln x)' = 1/x$, $(\sin x)' = \cos x$ and $(x^2)' = 2x$. Putting all of this together:

\begin{array}{ccc} \left( \frac{\ln x + \sin x}{x^2}\right)' &=& \frac{(1/x+\cos x)x^2-(\ln x + \sin x)(2x)}{(x^2)^2} \\ \\ &=& \frac{x+x^2\cos x - 2x\ln x - 2x\sin x}{x^4} \\ \\ &=& \frac{1}{x^3} + \frac{\cos x}{x^2}-2\frac{\ln x}{x^3} - 2\frac{\sin x}{x^3} \end{array}

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