## Finding Derivative Using The Definition vs. The Product Rule

Limits, differentiation, related rates, integration, trig integrals, etc.
simpleman
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### Finding Derivative Using The Definition vs. The Product Rule

My Text Book asked me to solve this problem using only the definition of the derivative. I found the correct answer using the definition of the derivative and then I wanted to test the product rule and the power rule to see if it would yield the same answer. To my frustration, I am getting a different answer. Can anyone see where I am mistaken?

Definition of Derivative: f'(x) = lim h-->0 (f(x+h)-f(x)) / h
Problem: find p'(x): p(x) = (3x+1)/(7x-6)
Work: (I would list all the algebra in detail but it's intensive and the answer is correct, per the back of the book)
Result: p'(x) = -25/(7x-6)2

The product rule: if F(x) = f(x)g(x), then: F'(x) = f(x)g'(x) + f'(x)g(x)
The power rule: if f(x) = xn, then: f'(x) = nxn-1
Problem: find p'(x): p(x) = (3x+1)/(7x-6)
Work:
for the product rule, let F(x) = p(x), f(x) = (3x+1) and g(x) = 1/(7x-6)
rewrite f(x) and g(x) as power expressions: let f(x) = (3x+1)1 and g(x) = (7x-6)-1
per the power rule, f'(x) = 1(3x+1)0 and g'(x) = -1(7x-6)-2
per the product rule: p'(x) =
= (3x+1)1 * -1(7x-6)-2 + 1(3x+1)0 * (7x-6)-1
= -(3x+1)/(7x-6)2 + 1/(7x-6)
= -(3x+1)*(7x-6) + (7x-6)2 / (7x-6)3
= -(3x+1) + (7x-6) / (7x-6)2
= 4x-7 / (7x-6)2
Result = 4x-7 / (7x-6)2

This result is clearly different than the answer found using the definition of the derivative. Please help! Thanks...

stapel_eliz
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for the product rule, let F(x) = p(x), f(x) = (3x+1) and g(x) = 1/(7x-6)
rewrite f(x) and g(x) as power expressions: let f(x) = (3x+1)1 and g(x) = (7x-6)-1
per the power rule, f'(x) = 1(3x+1)0 and g'(x) = -1(7x-6)-2
Don't you need to account for the derivatives of the insides of the parentheticals? For instance, in your derivative of f = (3x + 1)1, don't you need to differentiate the 3x + 1?

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