Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$g(x) = \int_{2x}^{6x} \frac{u+2}{u-4}du $$

For finding the $ g'(x)$, would I require to find first the derivative of $\frac{u+2}{u-4}$

then Replace the $u$ with 6x and 2x and add them ? (the 2x would have to flip so the whole term is negative)

If the previous statement is true would the final showdown be the following: $$ \frac{6}{(2x-4)^2} - \frac{6}{(6x-4)^2}$$

share|improve this question
1  
I think you do not need to differentiate the integrand. Hint: if $f(x) = \int_0^x \frac{u+2}{u-4}du$, then $f'(x) = \frac{x+2}{x-4}$, so try to express $g$ in terms of $f$. –  Lior B-S Jan 28 '13 at 7:06
    
Are you saying: $$6\frac{6x+2}{6x-4}+2\frac{2x+2}{2x-4} $$?? –  Reza M. Jan 28 '13 at 7:09
1  
Almost right, but minus, not plus. –  André Nicolas Jan 28 '13 at 7:12

2 Answers 2

up vote 3 down vote accepted

Let $f(u)=\frac{u+2}{u-4}$, and let $F(u)$ be the antiderivative of $f(u)$. Then $$ g'(x)=\frac{d}{dx}\int_{2x}^{6x}f(u)du=\frac{d}{dx}\left(F(u)\bigg\vert_{2x}^{6x}\right)=\frac{d}{dx}[F(6x)-F(2x)]=6F'(6x)-2F'(2x) $$ But $F'(u)=f(u)$. So the above evaluates to $$ 6f(6x)-2f(2x)=6\frac{6x+2}{6x-4}-2\frac{2x+2}{2x-4}=\cdots. $$ In general, $$ \frac{d}{dx}\int_{a(x)}^{b(x)}f(u)du=f(b(x))\cdot b'(x)-f(a(x))\cdot a'(x). $$

share|improve this answer
    
Oh! I understand, the dx(derivative) cancels out the integral(Anti-Derivative) So it stays the same. Then all that is required is to plug and play the correct areas. –  Reza M. Jan 28 '13 at 7:15
    
@RezaM. Taking the derivative more or less "undoes" the anti-derivative, but in general you have to be careful about the bounds of integration. –  BDub Jan 28 '13 at 7:17

According to Leibniz integral rule $$g'(x)=f(6x)\times 6-f(2x)\times 2$$ wherein $f(t)=\large \frac{t+2}{t-4}$.

share|improve this answer
    
+1 I just made your fraction a teeny bit bigger! :-) –  amWhy Jan 28 '13 at 16:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.