## Calculus: Early Transcendentals (2nd Edition)

$(a)$. $$\frac{dy}{dx}=e^{-y}$$ $(b)$. The slope of the tangent line is $m=\frac{1}{2}$
$(a)$. Use implicit differentiation to find $\frac{dy}{dx}$ for $x=e^y$ Taking the derivative implicitly we get: $$1=e^y\frac{dy}{dx}$$ Solve for $\frac{dy}{dx}$ $$\frac{dy}{dx}=\frac{1}{e^y}=e^{-y}$$ $(b)$. Finding slope of tangent line at the point $(2,\ln2)$ for the above function. We do this by plugging in the point $(2,\ln2)$ into the derivative from part $(a)$. Hence: $$\frac{dy}{dx}=e^{-(\ln2)}=e^{(\ln2^{-1})}=2^{-1}=\frac{1}{2}$$