Finding the constant in the derivative of (2x+1)^11(5x-1)^9

[neutron star]

Homework Statement

f(x)=(2x+1)^11 (5x-1)^9

Homework Equations

The Attempt at a Solution

f'(x)=[(2x+1)^10 (5x-1)^8] * (200x+23)
Where does the +23 come from, I don't get it. I get the rest.

 

[Mark44]

Your function and it's derivative are difficult to comprehend, as you have an error in your LaTeX.
I believe this is what you meant:
$$f(x)=(2x+1)^{11} (5x-1)^9$$
$$f'(x)=[(2x+1)^{10} (5x-1)^8] * (200x+23)$$


If you have an exponent that consists of more than one character, you have to put braces, {}, around it.

Using the product rule, f'(x) = 11*2*(2x + 1)¹⁰*(5x - 1)⁹ + (2x + 1)¹¹*9*5*(5x - 1)⁸
= 22*(2x + 1)¹⁰*(5x - 1)⁹ + 45*(2x + 1)¹¹*(5x - 1)⁸
= (2x + 1)¹⁰*(5x - 1)⁸[22(5x - 1) + 45(2x + 1)]
= (2x + 1)¹⁰*(5x - 1)⁸[110x - 22 + 90x + 45]
= (2x + 1)¹⁰*(5x - 1)⁸[200x + 23]

 

[rasmhop]

EDIT: Way too late. Disregard this, Mark44 beat me to it.

Homework Statement

f(x)=(2x+1)^11 (5x-1)^9

Homework Equations

The Attempt at a Solution

f'(x)=[(2x+1)^10 (5x-1)^8] * (200x+23)
Where does the +23 come from, I don't get it. I get the rest.

You use the product rule to get:
$$f'(x) = (5x-1)^9\frac{d}{dx}(2x+1)^{11} + (2x+1)^{11} \frac{d}{dx}(5x-1)^9$$
Use the chain-rule to calculate these two derivatives. Then with a little bit of rearranging you should get the desired result. As to where the 23 comes from it occurs naturally when you do the calculation, or a bit more concretely:
$$(5x-1)11(2) + (2x+1)9(5) = 110x+90x+45-22 = 200x+23$$
which you shall find when you do the calculations.

I don't see how you can get the rest of the answer unless you made a mistake. Try showing how you got the rest, but not the +23.