# Math Help - Equation of the tangent

1. ## Equation of the tangent

Determine the equation of the tangent to the curve f(x)=x*sqrt(1-x^2) at x=1/2 ???

What I have done:
y=sqrt3/4
((1/2),(sqrt3/4))

I am having problem finding the derivative f'(1/2). I know that f'(1/2) = 1/sqrt3

Can anyone help me by showing me how they have done f'(1/2) step by step

2. Originally Posted by sf1903
Determine the equation of the tangent to the curve f(x)=x*sqrt(1-x^2) at x=1/2 ???

What I have done:

$y=\frac{\sqrt{3}}{4}$

$(x,y)=\left(\frac{1}{2},\frac{\sqrt{3}}{4}\right)$

I am having problem finding the derivative f'(1/2). I know that f'(1/2) = 1/sqrt3

Can anyone help me by showing me how they have done f'(1/2) step by step
Use the product rule of differentiation..

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

Use the chain rule to differentiate $\sqrt{1-x^2}$

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

Now substitute x=0.5 to get

$\displaystyle\ f'(0.5)=\frac{1}{\sqrt{3}}$

3. It might help you to write f(x) as $f(x)= x(1- x^2)^{1/2}$. You will need to use the power rule, the product rule, and the chain rule.