If I have f(x)=(x^4)ln(x^2+2), how do I find the derivative? My thought: Take the 1st x der. of 2nd + 2nd x der. of 1st? (x^4) x (1/x^2+2) + ln(x^2+2) x (4x^3) Which would then equal ????
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Originally Posted by tradar If I have f(x)=(x^4)ln(x^2+2), how do I find the derivative? My thought: Take the 1st x der. of 2nd + 2nd x der. of 1st? (x^4) x (1/x^2+2) + ln(x^2+2) x (4x^3) Which would then equal ???? You are close. You just forgot to use the chain rule. $\displaystyle \frac{d}{dx} \ln(x^2+2)=\frac{2x}{x^2+2}$ Besides that you are good. What do you mean "which would equal"?
Originally Posted by tradar If I have f(x)=(x^4)ln(x^2+2), how do I find the derivative? My thought: Take the 1st x der. of 2nd + 2nd x der. of 1st? (x^4) x (1/x^2+2) + ln(x^2+2) x (4x^3) Which would then equal ???? $\displaystyle x^4 \cdot \frac{2x}{x^2+2} + 4x^3 \cdot \ln(x^2+2)$
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