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# Cal 1: Chain Rule question?

How do you do chain rule when you have a quotient?

y=(x+2/x-1)^7

and what if the exponent was only on either the numerator or the denominator? what would i do

Thanks!

### 4 Answers

- Anonymous9 years agoFavorite Answer
u = (x+2)/(x-1) ---> du/dx = -3/(x-1)^2

now use the chain rule

- Eclipse-girlLv 79 years ago
When you use the product rule, in the numerator you take the derivates of the numerator and the denominator. That is when you will end up using the chain rule

f(x) = g(x) / p(x)

f'(x) = (g'(x) p(x) - g(x) p'(x))/ (p(x)^2

g(x) = (x + 2)^7 g'(x) = 7(x + 2)^6 since the derivate of (x + 2) = 1 there is no more to do

p(x) = (x - 1)^7 p'(x) = 7(x -1)^6 since the derivative of (x - 1) = 1 there is no more to do

f'(x) = (((7(x + 2)^6)(x - 1)^ 7) - ((x + 2)^7)(7(x - 1)^6))) / (x -1)^14

Then you can simplify this.

There is 7(x+2)^6(x-1)^6 that is common to both terms in the numerator

After simplification the term in the denominator should be to the 8th power

A second way would be do

f(x) = g(x)^7 f'(x) = 7 g(x)^ 6 g'(x)

f'(x) = 7 (x + 2/x - 1) ^6 (the derivative of (x +2/x - 1) using the quotient rule)

Both will yield the same answer.

- Anonymous9 years ago
I'm pretty sure that you meant the first, but in case you didn't, i included the derivative of two functions that you may have intended to inquire about:

The derivative of (x+(2/x)-1)^7 = 7(1-(2/x^2))(x+(2/x)-1)^6

The derivative of (x+2)/(x-1)^7 = -(6x+15)/(x-1)^8

The derivative of ((x+2)/(x-1))^7 = -((21x+42)^6)/((x-1)^8)

Remember that the chain rule is:

if h(x)=f(g(x)), then h'(x)=f'(g(x))g'(x)

Source(s): http://www.calcreview.com/Derivatives-ab http://www.nyu.edu/academics/open-education/course... - Anonymous9 years ago
The derivative for A/B is ( (A' * B) - (A * B') )/ (B^2)

you can use that in the chain rule or for the second case