# finding the derivative

• Oct 11th 2011, 03:34 PM
habibixox
finding the derivative
Let http://webwork.rutgers.edu/webwork2_...da37dc1f31.png.

I got

[(-28x^3)(sqrtx) - (7x^4)(1/2(sqrtx))] - [(3x^2)(sqrtx)+(x^3)(1/2sqrtx)]/[(x^3sqrtx)^2]

what did i do wrong?
thanks
• Oct 11th 2011, 04:17 PM
Soroban
Re: finding the derivative
Hello, habibixox!

Quote:

$\displaystyle \text{Differentiate: }\:f(x) \;=\;-7x^4\sqrt{x} + \frac{7}{x^3\sqrt{x}}$

HEY! . . . Simplify first!

$\displaystyle f(x) \;=\;-7x^4\sqrt{x} + \frac{7}{x^3\sqrt{x}} \;=\;-7x^4x^{\frac{1}{2}} + \frac{7}{x^3x^{\frac{1}{2}}} \;=\;-7x^{\frac{9}{2}} + \frac{7}{x^{\frac{7}{2}}}$

. . $\displaystyle f(x) \;=\;-7x^{\frac{9}{2}} + 7x^{-\frac{7}{2}}$

Therefore: .$\displaystyle f'(x) \;=\;-\tfrac{63}{2}x^{\frac{7}{2}} - \tfrac{49}{2}x^{-\frac{9}{2}} \;=\;-\tfrac{7}{2}x^{-\frac{9}{2}}\left(9x^8 + 7\right)$

• Oct 11th 2011, 04:24 PM
habibixox
Re: finding the derivative
wow that's so much neater xD

thank you so much!!!
• Oct 11th 2011, 05:37 PM
Prove It
Re: finding the derivative
Quote:

Originally Posted by Soroban
Hello, habibixox!

HEY! . . . Simplify first!

$\displaystyle f(x) \;=\;-7x^4\sqrt{x} + \frac{7}{x^3\sqrt{x}} \;=\;-7x^4x^{\frac{1}{2}} + \frac{7}{x^3x^{\frac{1}{2}}} \;=\;-7x^{\frac{9}{2}} + \frac{7}{x^{\frac{7}{2}}}$

. . $\displaystyle f(x) \;=\;-7x^{\frac{9}{2}} + 7x^{-\frac{7}{2}}$

Therefore: .$\displaystyle f'(x) \;=\;-\tfrac{63}{2}x^{\frac{7}{2}} - \tfrac{49}{2}x^{-\frac{9}{2}} \;=\;-\tfrac{7}{2}x^{-\frac{9}{2}}\left(9x^8 + 7\right)$

It's also common practice to write your answer in the same form as the question, i.e. since the question was given in square root form, the answer should be given in square root form...