# Thread: finding the derivative

1. ## finding the derivative

Let .

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

2. ## Re: finding the derivative

Hello, habibixox!

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

HEY! . . . Simplify first!

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

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

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

3. ## Re: finding the derivative

wow that's so much neater xD

thank you so much!!!

4. ## Re: finding the derivative

Originally Posted by Soroban
Hello, habibixox!

HEY! . . . Simplify first!

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

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

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

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...