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$$g(x) = \int_{2x}^{6x} \frac{u+2}{u-4}du $$ 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: $$ \frac{6}{(2x-4)^2} - \frac{6}{(6x-4)^2}$$ |
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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, $$ \frac{d}{dx}\int_{a(x)}^{b(x)}f(u)du=f(b(x))\cdot b'(x)-f(a(x))\cdot a'(x). $$ |
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According to Leibniz integral rule $$g'(x)=f(6x)\times 6-f(2x)\times 2$$ wherein $f(t)=\large \frac{t+2}{t-4}$. |
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