Question

y=sqrt(1+4sinx) (0,1) find the tangent to the curve at given point

Answer 1

\[\huge y=\sqrt{1+4sinx}=(1+4sinx)^{\frac{1}{2}} \]
\[\huge y'=\frac{1}{2}(1+4sinx)^{-\frac{1}{2}}\cdot (4cosx) \]

Answer 2

now evaluate the derivative at x=0 to find the slope of the tangent line.

Answer 3

so do you plug it in?

Answer 4

yes...

Answer 5

\[\huge y'=\frac{1}{2}(1+4sinx)^{-\frac{1}{2}}\cdot (4cosx) \]
\[\huge y'|_{x=0}=\frac{1}{2}(1+4sin(0))^{-\frac{1}{2}}\cdot (4cos(0)) \]
\[\huge y'|_{x=0}=\frac{1}{2}(1+(0))^{-\frac{1}{2}}\cdot 4(1) \]
\[\huge y'|_{x=0}=\frac{1}{2}\cdot 4 = 2\]

Answer 6

k thats what i got

Answer 7

y=2x+9

Answer 8

now that you have a slope, m=2, and a point, (0, 1), put this into the point-slope form:
\[\huge y-y_0=m(x-x_0) \]

Answer 9

\[\huge y-y_0=m(x-x_0) \]
\[\huge y-1=2(x-0) \]
\[\huge y-1=2x \]
\[\huge y=2x+1 \]

Answer 10

ohhh ok!