Question

Find the slope of the tangent line to the curve:

Answer 1

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

Answer 2

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
\[\sqrt{x}\]

Answer 3

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

Answer 4

oh yeah i got that

Answer 5

then we simply plug in 4 right

Answer 6

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

Answer 7

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

Answer 8

and that would be our slope right

Answer 9

yup!