1. ## Derivatives problem help.

How would I start for a problem like this?

Can anyone solve this problem with steps. I don't know what to do after using the quotient and product rule on the problem below.

Find the derivative of the function.
g(x) = (4 + 5x)^5(4 + x - x^2)^6

Another problem I am stuck on is the one below. How would I start the problem, can anyone show the steps to solve the problem?

Find the derivative of the function.
G(y)= $\displaystyle (y-2)^4/(y^2+4y)^9$

Thanks for any replies, it would be great to get a reply as soon as possible.

2. after using product rule for the first one, you should get:

$\displaystyle 5(4+5x)^4 \times (5) \times (4 + x - x^2)^6 + 6(4 + x - x^2)^5 \times (1-2x) \times (4+5x)^5$

$\displaystyle = 25(4+5x)^4(4 + x - x^2)^6+6(1-2x)(4 + x - x^2)^5(4+5x)^5$
Further simplification can be done.

What problem did you have in the second one? Its a straight application of Quotient Rule plus chain rule... SHow your work if you have any problem..

3. How would I simplify the first product rule further, would I multiply some things out?

For the second problem, I get stuck after this step.

(y^2 + 4y)^9· 4(y − 2)^3· 1 − (y − 2)^4· 9(y^2 + 4y)^8(2y + 4)/
[(y^2 + 4y)^9]^2

4. $\displaystyle = 25(4+5x)^4(4 + x - x^2)^6+6(1-2x)(4 + x - x^2)^5(4+5x)^5$

Take $\displaystyle (4+5x)^4 \mbox{and} (4 + x - x^2)^5$ as a common factor:

$\displaystyle = (4+5x)^4(4 + x - x^2)^5[25(4 + x - x^2)+6(1-2x)(4+5x)]$

For the second one,

your notations are hard to understand, but I think you are close to the answer. It is:

$\displaystyle \displaystyle{\frac{[(y^2+4y)^9 \times 4(y-2)^3] - [(y-2)^4 \times 9(y^2+4y)^8 \times(2y+4)]}{(y^2+ 4y)^{18}}}$

5. My book did a sample problem in a different way, how would I get the answer into this form?

Here are the steps. The way it did the problem confused me.

6. $\displaystyle \displaystyle{\frac{[(y^2+4y)^9 \times 4(y-2)^3] - [(y-2)^4 \times 9(y^2+4y)^8 \times(2y+4)]}{(y^2+ 4y)^{18}}}$

In the numerator take $\displaystyle (y^2+4y)^8 \mbox{and} (y-2)^3$ as common terms:

$\displaystyle = \displaystyle{\frac{(y^2+4y)^8(y-2)^3[4(y^2+4y)-9(y-2)(2y+4)]}{(y^2+ 4y)^{18}}}$

Rest of the multiplication and cancellation is for you...

7. Thanks for the help, can you simplify the [25 times etc stuff] in the first problem to get a more simplified answer or is the answer simplified enough?

8. $\displaystyle [25(4 + x - x^2)+6(1-2x)(4+5x)]$

You can multiply and add up like terms here..

9. Would you approach problems like y=arcsin(5x) the same way with the product rule?

10. Product rule????

If you are talking about differentiation here, you MUST have heard of DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS.

11. So it's a special way of doing the derivative?

I've heard of it.

12. Originally Posted by ioke09
So it's a special way of doing the derivative?

I've heard of it.

What class are you taking? Isnt it calculus I?? Anyways, this link might help

13. Find y' by implicit differentiation.
cos(x) + √y = 6

I'm not sure if I did this correctly, am I on the right track?

I have -sinx+ 1/2y^-1/2 y'=0 so far.