- #1

mooha

- 5

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**Taking the derivative of a polynomial fraction??**

b]1. Homework Statement [/b]

Ok, so the question wants me to differentiate f(x)= (x)/(x+1). We are supposed to use the definition of the derivative f'(x)= (limit as h->0) [f(x+h)-f(x)]/(h). We also have learned the power rule. I did the formula and had some issues, so I tried to check my answer with the power rule. However they didn't come out the same. I'm not sure where I made the mistake, the formula or the power rule!

**2. Homework Equations [/b**

f'(x)= (limit as h->0) [f(x+h)-f(x)]/(h)

Here's my work:

f'(x)= (lim h->0) [(x+h)/(x+h+1) - x/(x-1)] / h

=(lim h->0) [(x+h)/(x+h+1) - x/(x-1)]/h * (x+h+1)(x+1)

= (lim h->0) [(x+h)(x+1) - (x)(x+h+1)] / (h)(x+h+1)(x+1)

= (lim h->0) (x^2+x+hx+h-x^2-xh-x)/(xh+h^2x+hx^2+h+h^2+hx)

= (lim h->0) h/(h)(x+hx+x^2+h+x)

=(lim h->0) 1/x^2+2x+hx+h

f'(x)= 1/x^2+2x

Power Rule:

y'= (x)/(x+1)

y'= (x)*(x+1)^-1

y'= -(x+1)^-2

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

Thank you so much!!

f'(x)= (limit as h->0) [f(x+h)-f(x)]/(h)

## The Attempt at a Solution

Here's my work:

f'(x)= (lim h->0) [(x+h)/(x+h+1) - x/(x-1)] / h

=(lim h->0) [(x+h)/(x+h+1) - x/(x-1)]/h * (x+h+1)(x+1)

= (lim h->0) [(x+h)(x+1) - (x)(x+h+1)] / (h)(x+h+1)(x+1)

= (lim h->0) (x^2+x+hx+h-x^2-xh-x)/(xh+h^2x+hx^2+h+h^2+hx)

= (lim h->0) h/(h)(x+hx+x^2+h+x)

=(lim h->0) 1/x^2+2x+hx+h

f'(x)= 1/x^2+2x

Power Rule:

y'= (x)/(x+1)

y'= (x)*(x+1)^-1

y'= -(x+1)^-2

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

Thank you so much!!