# Thread: find slope of tangent line at x=a

1. ## find slope of tangent line at x=a

for y=1/(sqrt(x)) find slope of tangent line at the point where x=a

then, find the equation of the tangent line at (4,1/2)

2. Take the first derivative and replace x with a to find your slope at point a.

For the second problem do this

$f'(4)= ?$

when you find out what the slope is

plug it into

$(y_2-y_1) = m(x_2-x_1)$

m being the slope

3. Originally Posted by mono1357
for y=1/(sqrt(x)) find slope of tangent line at the point where x=a

then, find the equation of the tangent line at (4,1/2)
The slope of the tangent line at $x=a$ is just $\left.\frac{\,dy}{\,dx}\right|_{x=a}$

Thus, $y=x^{-\frac{1}{2}}\implies \frac{\,dy}{\,dx}=-\tfrac{1}{2}x^{-\frac{3}{2}}$.

Thus, at $x=a$, $\left.\frac{\,dy}{\,dx}\right|_{x=a}=-\tfrac{1}{2}a^{-\frac{3}{2}}$

Now, let's find the equation of the tangent line:

$y-f(a)=f'(a)(x-a)$

Since $a=4,~f(a)=\tfrac{1}{2},~\text{and }f'(a)=-\tfrac{1}{16}$, we now see that $y=-\tfrac{1}{16}(x-4)+\tfrac{1}{2}\implies {\color{red}y=-\tfrac{1}{16}x+\tfrac{3}{4}}$

Does this make sense?

--Chris