how to find the equation of a tangent line y= sqrt 1+4 sinx at point (0,1)

Question

anonymous · Answer

The tangent line $y$ of a function $f(x)$ at the point $(a,f(a))$ is: $$
y-f(a)=f'(a)(x-a)
$$

anonymous · Answer

In this case $a=0,f(a)=1$.
You need to find the derivative of $f(x)=\sqrt{1+4\sin(x)}$

anonymous · Answer

okay, so far i have y'= 1/2 (1+4sinx)^-1/2 (cosx) but idk if the cosx is right or what to do then.

anonymous · Answer

So you want help finding the derivative?

anonymous · Answer

yah

anonymous · Answer

I would do implicit differentiation: $$
\begin{array}{}
y&=&\sqrt{1+4\sin(x)}\
y^2&=&1+4\sin(x)\
y^2-1&=&4\sin(x)\
2yy'&=&4\cos(x)\
y' &=&\frac{4\cos(x)}{2y}\
y' &=&\frac{4\cos(x)}{2\sqrt{1+4\sin(x)}}
\end{array}
$$

anonymous · Answer

You can also use chain rule: $$
y=(f\circ g)\
f=\sqrt{x}\
g=1+4\sin(x)\
$$$$
f'=\frac{1}{2\sqrt{x}}\
g'=4\cos(x)\
y'=f(g)\cdot g'=\frac{4\cos(x)}{2\sqrt{1+4\sin(x)}}
$$

anonymous · Answer

@ansleyb Do you think you can finish the rest?

anonymous · Answer

yah thank you