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)

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- Sep 11th 2008, 08:46 PMmono1357find 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) - Sep 11th 2008, 08:52 PM11rdc11
Take the first derivative and replace x with a to find your slope at point a.

For the second problem do this

$\displaystyle f'(4)= ?$

when you find out what the slope is

plug it into

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

m being the slope - Sep 11th 2008, 08:56 PMChris L T521
The slope of the tangent line at $\displaystyle x=a$ is just $\displaystyle \left.\frac{\,dy}{\,dx}\right|_{x=a}$

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

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

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

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

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

Does this make sense?

--Chris