OpenStudy (anonymous):

Find the slope of the tangent line to the curve:

6 years ago
OpenStudy (anonymous):

$y=\int\limits_{0}^{\sqrt{x}} e ^{-t^2} dt (x>0) at x=4$

6 years ago
OpenStudy (anonymous):

take the derivative using the chain rule. the derivative of the integral is the integrand, but this time the upper limit is $\sqrt{x}$ so you have to replace t by $\sqrt{x}$ and also multiply by the derivative of $\sqrt{x}$

6 years ago
OpenStudy (anonymous):

$y'=e^{-(\sqrt{x})^2}\times \frac{1}{2\sqrt{x}}$

6 years ago
OpenStudy (anonymous):

oh yeah i got that

6 years ago
OpenStudy (anonymous):

then we simply plug in 4 right

6 years ago
OpenStudy (anonymous):

good so we get $y'=e^{-x}\times \frac{1}{2\sqrt{x}}$

6 years ago
OpenStudy (anonymous):

plug in 4 get out $\frac{e^{-4}}{4}$

6 years ago
OpenStudy (anonymous):

and that would be our slope right

6 years ago
OpenStudy (anonymous):

yup!

6 years ago