1. ## product rule

Dervative of (3x+2)^4 (5x+2)^-5

2. Originally Posted by Nimmy
Dervative of (3x+2)^4 (5x+2)^-5
Use the product rule. Call $f(x) = (3x+2)^4$ and $g(x) = (5x+2)^{-5}$

Then $f'(x) = 4(3x+2)^3 \cdot 3 = 12(3x+2)^3$.

And $g'(x) = -5(5x+2)^{-6} \cdot 5 = -25(5x+2)^{-6}$.

Thus the derivative of $(3x+2)^4 (5x+2)^{-5}$ is $f'(x)g(x) + f(x)g'(x)$

= $12(3x+2)^3 \cdot (5x+2)^{-5} + (3x+2)^4 \cdot -25(5x+2)^{-6}$

= $12(3x+2)^3(5x+2)^{-5} - 25(3x+2)^4(5x+2)^{-6}$

-Dan

3. Originally Posted by topsquark
Use the product rule. Call $f(x) = (3x+2)^4$ and $g(x) = (5x+2)^{-5}$

Then $f'(x) = 4(3x+2)^3 \cdot 3 = 12(3x+2)^3$.

And $g'(x) = -5(5x+2)^{-6} \cdot 5 = -25(5x+2)^{-6}$.

Thus the derivative of $(3x+2)^4 (5x+2)^{-5}$ is $f'(x)g(x) + f(x)g'(x)$

= $12(3x+2)^3 \cdot (5x+2)^{-5} + (3x+2)^4 \cdot -25(5x+2)^{-6}$

= $12(3x+2)^3(5x+2)^{-5} - 25(3x+2)^4(5x+2)^{-6}$

-Dan
If you need that in fraction form:

(Continuing)

= $\frac{12(3x+2)^3}{(5x+2)^5} - \frac{25(3x+2)^4}{(5x+2)^6}$

= $\frac{12(3x+2)^3}{(5x+2)^5} \cdot \frac{5x+2}{5x+2} - \frac{25(3x+2)^4}{(5x+2)^6}$

= $\frac{12(3x+2)^3(5x+2) - 25(3x+2)^4}{(5x+2)^6}$

= $\frac{(3x+2)^3(12(5x+2) - 25(3x+2))}{(5x+2)^6}$

= $\frac{(3x+2)^3(60x + 24 - 75x - 50)}{(5x+2)^6}$

= $\frac{(3x+2)^3(-15x - 26)}{(5x+2)^6}$

= $-\frac{(3x+2)^3(15x + 26)}{(5x+2)^6}$

-Dan

4. Originally Posted by Nimmy
Dervative of (3x+2)^4 (5x+2)^-5
Or you can use the quotient rule. Let $f(x) = (3x+2)^4$ and $g(x) = (5x+2)^5$.

Then $f'(x) = 4(3x+2)^3 \cdot 3 = 12(3x+2)^3$.

And $g'(x) = 5(5x+2)^4 \cdot 5 = 25(5x+2)^4$.

Then the derivative of $\frac{(3x+2)^4}{(5x+2)^5}$ is $\frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$

= $\frac{12(3x+2)^3 \cdot (5x+2)^5 - (3x+2)^4 \cdot 25(5x+2)^4}{ \left ( (5x+2)^5 \right ) ^2}$

= $\frac{(5x+2)^4(12(3x+2)^3 \cdot (5x+2) - 25(3x+2)^4)}{(5x+2)^10}$

= $\frac{12(3x+2)^3(5x+2) - 25(3x+2)^4}{(5x+2)^6}$

which is the same as the third line in my second post.

-Dan