Ask your own question, for FREE!
Mathematics
OpenStudy (anonymous):

Find the slope of the tangent line to the curve:

OpenStudy (anonymous):

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

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 \[\sqrt{x}\]

OpenStudy (anonymous):

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

OpenStudy (anonymous):

oh yeah i got that

OpenStudy (anonymous):

then we simply plug in 4 right

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

and that would be our slope right

OpenStudy (anonymous):

yup!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!