## Calculus: Early Transcendentals 8th Edition

The equation of the tangent line is $y=2x+1$
$y=\sqrt{1+4\sin x}$ $,$ $(0,1)$ Rewrite the expression using a rational exponent: $y=(1+4\sin x)^{1/2}$ Apply the chain rule to evaluate the derivative of the given expression: $y'=\dfrac{1}{2}(1+4\sin x)^{-1/2}(1+4\sin x)'=...$ Evaluate the indicated derivative and simplify: $...=\dfrac{1}{2}(1+4\sin x)^{-1/2}(4\cos x)=\dfrac{2\cos x}{\sqrt{1+4\sin x}}$ Substitute $x$ by $0$ in the derivative found to obtain the slope of the tangent line at the given point: $m=\dfrac{2\cos0}{\sqrt{1+4\sin0}}=\dfrac{2(1)}{\sqrt{1+4(0)}}=\dfrac{2}{\sqrt{1}}=2$ Both the slope of the tangent line and a point through which it passes are known. Use the point-slope formula of the equation of a line, which is $y-y_{1}=m(x-x_{1})$, to obtain the equation of the tangent line at the given point: $y-1=2(x-0)$ $y-1=2x$ $y=2x+1$ The equation of the tangent line is $y=2x+1$