# Thread: Derivative of a curve

1. ## Derivative of a curve

Find $\displaystyle \frac{dy}{dx}$ of the curve $\displaystyle y=\sqrt{4x-7}-\frac{2}{3}(x-4)$ and show that $\displaystyle \frac{d^2y}{dx^2} = \frac{-4}{(4x-7)\sqrt{4x-7}}$

2. Originally Posted by Punch
Find $\displaystyle \frac{dy}{dx}$ of the curve $\displaystyle y=\sqrt{4x-7}-\frac{2}{3}(x-4)$ and show that $\displaystyle \frac{d^2y}{dx^2} = \frac{-4}{(4x-7)\sqrt{4x-7}}$
What have you tried and where do you get stuck?

3. Originally Posted by mr fantastic
What have you tried and where do you get stuck?
I am stucked at the part where I am asked to show.
My final answer to $\displaystyle \frac{dy}{dx}$ is $\displaystyle \frac{6-2(4x-7)}{3(4x-7)}$

4. Originally Posted by Punch
I am stucked at the part where I am asked to show.
My final answer to $\displaystyle \frac{dy}{dx}$ is $\displaystyle \frac{6-2(4x-7)}{3(4x-7)}$
That's not correct. If you show your working your mistakes can be pointed out.

5. From $\displaystyle y=\sqrt{4x-7}-\frac{2}{3}(x-4)$

$\displaystyle \frac{dy}{dx}=\frac{1}{2}(4x-7)^{\frac{-1}{2}}(4)-(\frac{2}{3}(1))$

$\displaystyle =\frac{2}{(4x-7)^{-0.5}}-\frac{2}{3}$

$\displaystyle =\frac{6-2(4x-7))^{-0.5}}{3(4x-7))^{-0.5}}$

6. Originally Posted by Punch
From $\displaystyle y=\sqrt{4x-7}-\frac{2}{3}(x-4)$

$\displaystyle \frac{dy}{dx}=\frac{1}{2}(4x-7)^{\frac{-1}{2}}(4)-(\frac{2}{3}(1))$

$\displaystyle =\frac{2}{(4x-7)^{-0.5}}-\frac{2}{3}$

$\displaystyle =\frac{6-2(4x-7))^{-0.5}}{3(4x-7))^{-0.5}}$
First, this is the not the result you posted earlier.

Secondly, it should be $\displaystyle \frac{2}{(4x-7)^{0.5}}-\frac{2}{3}$ not $\displaystyle \frac{2}{(4x-7)^{{\color{red}-}0.5}}-\frac{2}{3}$.

And thirdly, I'd leave it as $\displaystyle =\frac{2}{(4x-7)^{0.5}}-\frac{2}{3}$ because this is easier to differentiate (to get the second derivative) than the final form you have given.

7. oh sorry, I made many typo errors... thanks for correcting.
I am still stucked at showing, glad to know that I am on the right path on getting derivative (thanks you).