How do I find f'(x) for f(x)=7(x^(4/5))-8(x^(6/7))?

I got as far as 5.6(x^(-1/5))-6.857(x^(-1/7)) but this does not appear to be either the final form of the answer or the correct answer according to the system I do homework on. I'm confused about the power rules, and would appreciate any help!

1 Answer
Feb 24, 2018

f^{'}(x)\ \ =\ \ \frac{28}{5x^{\frac{1}{5}}}-\frac{48}{7x^{\frac{1}{7}}}

Explanation:

f(x) = 7x^{\frac{4}{5}}-8x^{\frac{6}{7}

f^'(x) = \frac{d}{dx}(7x^{\frac{4}{5}}-8x^{\frac{6}{7}})

\text{Apply the Sum/Difference of derivative Rule}:

\quad (f(x)\pm g(x))^'=f^'(x)\pm g^'(x)

=\frac{d}{dx}(7x^{\frac{4}{5}})-\frac{d}{dx}(8x^{\frac{6}{7}})


Take the constant out \ \ \ \ (a\cdot f(x))^'=a\cdot f^'(x)\ \ \ \ and apply the power rule \ \ \ \ \quad \frac{d}{dx}(x^a)=a\cdot x^{a-1}\ \ \ \ to get:

=7\frac{d}{dx}(x^{\frac{4}{5}})-8\frac{d}{dx}(x^{\frac{6}{7}})

=7\cdot \frac{4}{5}x^{\frac{4}{5}-1}-8\cdot \frac{6}{7}x^{\frac{6}{7}-1}

Simplify the exponents and then rewrite the terms by using the exponent rule #\ \ \# a^{-b}=\frac{1}{a^b}

=\frac{28}{5}x^{-\frac{1}{5}}\ \ -\ \ \frac{48}{7}x^{-\frac{1}{7}}

=\frac{28}{5x^{\frac{1}{5}}}-\frac{48}{7x^{\frac{1}{7}}}


That's it!