Question

find the derivative of (3x-2)^10 (5x^2 -x +1)^12

Answer 1

Apply product rule, power rule, chain rule.

Answer 2

\[
\frac{d}{dx} (3x-2)^{10}(5x^2 -x +1)^{12} = \\
(3x-2)^{10} * \frac{d}{dx} (5x^2 -x +1)^{12} +
\frac{d}{dx} (3x-2)^{10} * (5x^2 -x +1)^{12} = ?
\]

Answer 3

Ok, thank you. I'll give it a try.

Answer 4

You are welcome.

Answer 5

Hi, I've done this problem multiple times now, and I keep on getting the wrong answer. The answer to the problem is: \[6(3x-2)^2(5x^2-x+1)^11(85x^2-51x+9)\]
Can someone please show me how to get this answer???

Answer 6

\[
\frac{d}{dx} (3x-2)^{10}(5x^2 -x +1)^{12} = \\
(3x-2)^{10} * \frac{d}{dx} (5x^2 -x +1)^{12} +
\frac{d}{dx} (3x-2)^{10} * (5x^2 -x +1)^{12} = \\
(3x-2)^{10} * 12 * (5x^2 -x +1)^{11} * (10x - 1) +
10(3x-2)^{9} * 3 * (5x^2 -x +1)^{12} = \\
(3x-2)^9 * (5x^2 -x +1)^{11} * 6 * ~\{~(3x-2)*2*(10x-1) + 5 * (5x^2 -x +1)~\} = \\
6(3x-2)^9 (5x^2 -x +1)^{11} \{~ 2(30x^2-3x-20x+2)+25x^2-5x+5 ~\} = \\
6(3x-2)^9 (5x^2 -x +1)^{11} \{~ 60x^2-6x-40x+4+25x^2-5x+5 ~\} = \\
6(3x-2)^9 (5x^2 -x +1)^{11} (85x^2-51x+9 )
\]

Answer 7

You are probably making a mistake in applying the power rule and then the chain rule.
For example, to find the derivative of \(\frac{d}{dx} (5x^2 -x +1)^{12} \), the first step is the power rule. The derivative of some_quantity raised to n is n times some-quantity raised to (n-1). That is the power rule part. Then comes the chain rule which is: times the derivative of some-quantity.
\[
\frac{d}{dx}(\text{some_quantity})^n = n * (\text{some_quantity})^{n-1} *
\frac{d}{dx}(\text{some_quantity}) \\
\frac{d}{dx} (5x^2 -x +1)^{12} =
12 * (5x^2 -x +1)^{11} * \frac{d}{dx} (5x^2 -x +1) = \\
12 * (5x^2 -x +1)^{11} * (10x - 1)
\]Similarly,
\[
\frac{d}{dx} (3x-2)^{10}= 10 * (3x-2)^9 * \frac{d}{dx} (3x-2) = \\
10 * (3x-2)^9 * 3 = 30(3x-2)^9
\]

Answer 8

That was extremely helpful! Thank you! I really appreciate the time that you put to write all of that down.

Answer 9

You are very welcome.

Answer 10

A Mathematica 9 solution and a detailed simplification from WolframAlfa is attached.