Answer
The equation of the tangent line is $y=2x+1$
Work Step by Step
$y=\sqrt{1+4\sin x}$ $,$ $(0,1)$
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 $y=2x+1$
