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
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