1. ## Differentiation - Gradient equals zero

Calculate the coordinates at which the gradient is zero for the curve $\displaystyle y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}$

Attemped solution

$\displaystyle Let u=\sqrt{1-x}$ and $\displaystyle v=\sqrt{x^2+3}$

$\displaystyle \frac{du}{dx}=\frac{1}{2\sqrt{1-x}}$

$\displaystyle \frac{dv}{dx}=\frac{x}{\sqrt{x^2+3}}$

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

$\displaystyle \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0$

Stucked here!! need help in solving for x...

2. Originally Posted by Punch
Calculate the coordinates at which the gradient is zero for the curve y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}

Attemped solution

$\displaystyle Let u=\sqrt{1-x}$ and $\displaystyle v=\sqrt{x^2+3}$

$\displaystyle \frac{du}{dx}=\frac{1}{2\sqrt{1-x}}$

$\displaystyle \frac{dv}{dx}=\frac{x}{\sqrt{x^2+3}$

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

$\displaystyle \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0$

Stucked here!! need help in solving for x...

$\displaystyle y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}$

Taking the derivative and setting to 0...

$\displaystyle y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}\Rightarrow\frac{d}{dx}[\frac{\sqrt{1-x}}{\sqrt{x^2+3}}]=?$

3. Originally Posted by VonNemo19
$\displaystyle y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}$

Taking the derivative and setting to 0...

$\displaystyle y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}\Rightarrow\frac{d}{dx}[\frac{\sqrt{1-x}}{\sqrt{x^2+3}}]=?$
Please look at my attemped solution, I have already tried to differentiate. The problem lies in solving for x....

4. Originally Posted by Punch
Calculate the coordinates at which the gradient is zero for the curve $\displaystyle y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}$

Attemped solution

$\displaystyle Let u=\sqrt{1-x}$ and $\displaystyle v=\sqrt{x^2+3}$

$\displaystyle \frac{du}{dx}=\frac{1}{2\sqrt{1-x}}$

$\displaystyle \frac{dv}{dx}=\frac{x}{\sqrt{x^2+3}}$

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

$\displaystyle \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0$

Stucked here!! need help in solving for x...

Multiply both sides by $\displaystyle x^2+ 3$ so $\displaystyle \sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})=0$

$\displaystyle \sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})=\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})$

Multiply both sides by $\displaystyle 2\sqrt{1-x}\sqrt{x^2+ 3}$ and you get rid of the fractions and the square roots.

5. Originally Posted by Punch
Calculate the coordinates at which the gradient is zero for the curve $\displaystyle y=\frac{\sqrt{1-x}}{\sqrt{x^2+3}}$

Attemped solution

$\displaystyle Let u=\sqrt{1-x}$ and $\displaystyle v=\sqrt{x^2+3}$

$\displaystyle \frac{du}{dx}=\frac{1}{2\sqrt{1-x}}$

$\displaystyle \frac{dv}{dx}=\frac{x}{\sqrt{x^2+3}}$

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

$\displaystyle \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0$

Stucked here!! need help in solving for x...

Actually $\displaystyle \frac{du}{dx} = -\frac{1}{2\sqrt{1 - x}}$

The rest of the derivative looks correct.

When you solve for $\displaystyle x$, try putting everything over a common denominator first...

6. thanks!