$\displaystyle f(x)= x^\frac{6}{9}$ I need help with fractional exponents. is it still $\displaystyle nx^n-1$
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Yes , if y=x^n then y' = n x^(n-1). So, in your problem , y'=(6/9)(x^((6/9)-1))= (2/3)(x^(-1/3))=2/(3x^(1/3)).
Originally Posted by Zanderist $\displaystyle f(x)= x^\frac{6}{9}$ I need help with fractional exponents. is it still $\displaystyle nx^n-1$ first, note that $\displaystyle \frac{6}{9} = \frac{2}{3}$ $\displaystyle f(x) = x^{\frac{2}{3}}$ $\displaystyle f'(x) = \frac{2}{3} x^{\frac{2}{3} - 1} =$ ?
Originally Posted by skeeter first, note that $\displaystyle \frac{6}{9} = \frac{2}{3}$ $\displaystyle f(x) = x^{\frac{2}{3}}$ $\displaystyle f'(x) = \frac{2}{3} x^{\frac{2}{3} - 1} =$ ? $\displaystyle f'(x) = \frac{2}{3} x^{-\frac{1}{3}}$ I have another one. $\displaystyle 4\sqrt{t}+\frac{7}{\sqrt{t}}$
Originally Posted by Zanderist $\displaystyle 4\sqrt{t}+\frac{7}{\sqrt{t}}$ First rewrite the equation: $\displaystyle 4t^{\frac{1}{2}}+7t^{\frac{-1}{2}}$ Now follow your rules $\displaystyle nx^{n-1}$
Originally Posted by dani First rewrite the equation: $\displaystyle 4t^{\frac{1}{2}}+7t^{\frac{-1}{2}}$ Now follow your rules $\displaystyle nx^{n-1}$ So...this.. $\displaystyle 4/2t^{\frac{1}{2}-1}+7/2t^{\frac{-1}{2}-1}$ to this... $\displaystyle 2t^{-\frac{1}{2}}+\frac{7}{2}t^{-\frac{3}{2}}$
Okay let's take it up a level... $\displaystyle \frac{e^x}{(2-1x)}$ Quotient rule I believe. I just want to make sure about this $\displaystyle f(x)=e^x$ It's derivative is... $\displaystyle f '(x)= e^x \log e $
I have a problem I just can't stand. I take it to find B you need the second der. for A you just leave it as is. I just don't know what is the proper answer.
start a new problem with a new thread, otherwise this one can get very confusing. thank you.