f(x)= ln(5x^4-x)
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Originally Posted by av8or91 f(x)= ln(5x^4-x) Use the chain rule $\displaystyle \frac{d}{dx}ln[f(x)] = \frac{f'(x)}{f(x)}$ In your case $\displaystyle f(x)=5x^4-x$ which s pretty simple to differentiate
Originally Posted by av8or91 f(x)= ln(5x^4-x) The derivative of $\displaystyle y=\ln(f(x))$ is $\displaystyle y'=\frac{f'(x)}{f(x)}$. So what do you think?
oh ok I see it now I think. So its just the derivative of the top/ original function on the bottom. 20x^3-1 5x^4-x
Originally Posted by av8or91 oh ok I see it now I think. So its just the derivative of the top/ original function on the bottom. 20x^3-1 5x^4-x Aside from the syntax then yes $\displaystyle \frac{20x^3-1}{5x^4-x}$
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