## Calculus: Early Transcendentals 8th Edition

$$y=(\ln2 )x+1$$
Re-arranging the function into a differentiable form: $y=e^{(\ln2) x}$ The derivative of the curve is (using the chain rule): $y'=\frac{de^{(\ln2)x}}{d(\ln2)x} \times\frac{d(\ln2)x}{dx}$ $=e^{(\ln 2)x} \times \ln2$ $=\ln 2\times2^x$ Thus, the slope ($m$) of the tangent line at (0,1) would be: $m= y'(0)$ $=\ln 2 \times 2^{(0)}$ (plugging $x=0$ into equation $y'$) $=\ln2$ Thus, plugging in the given points (0,1) into the general point-slope form of a linear equation, an equation of the tangent line is: $y-1=\ln2(x-0)$ Further simplifying the equation gives: $y=(\ln2 )x+1$