The funciton is 3x/(x^2+4) I have no idea how to differentiate it. Step by step instructions would really help.
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Use the Quotient Rule.
What's the quotient rule? I know the product rule is f(x)g'(x)+f'(x)g(x)
Originally Posted by AFireInside What's the quotient rule? I know the product rule is f(x)g'(x)+f'(x)g(x) If you don't know the Quotient Rule, you can use the Product Rule if you rewrite the function as 3x(x^2 + 4)^(-1)
But what do i do with the -1 exponent?
Originally Posted by AFireInside But what do i do with the -1 exponent? You would need to use the chain rule to differentiate (x^2 + 4)^(-1)
But then the -1 becomes a -2 and it doesn't really get you anywhere
Originally Posted by AFireInside But then the -1 becomes a -2 and it doesn't really get you anywhere What's wrong with that?
Well using the chain rule (x^2+4)^-1 would become -(x^2+4)^-2 x (x^2+4). I don't see how that simplifies things.
No it doesn't... It's the derivative of the inside times the derivative of the outside.
Now I'm lost.
The inside is u = x^2 + 4. What is its derivative? The outside is u^(-1). What is its derivative?
Originally Posted by AFireInside Well using the chain rule (x^2+4)^-1 would become -(x^2+4)^-2 x (x^2+4). I don't see how that simplifies things. If $\displaystyle y(x) = \frac{{f(x)}}{{g(x)}}$ then $\displaystyle y'(x) = \frac{{f'(x)g(x) - f(x)g'(x)}}{{\left[ {g(x)} \right]^2 }}$. $\displaystyle f(x)=3x~\&~g(x)=x^2+4$
I thought chain rule was nf'(x) x f(x)^n-1 So it should be -1(2x)(x^+4)^-2
Correct, despite the typo, it should be -2x(x^2 + 4)^(-2). So now you can apply the product rule.
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