So I see you don't understand what the question is asking. I'm going to give a brief rundown of the steps for this problem before going into them so you get an idea of what's being asked.
1.) We will solve for the y-value at #x=0# so we get a full coordinate point.
2.) We are going to find the derivative of the function.
3.) we are going to solve for the derivative at #x=0#. This will give us the slope of the tangent line at that point.
4.) We will plug in the coordinate points into point slope form along with the calculated slope value from the derivative to get us an equation for the tangent line.
5.) Using this equation, we will plug in #x=.12# to get an approximate value of the function at that point.
1.) So when we plug 0 into the equation, we get #y=(4+sin(0))^.5#. #sin(0)=0#, so #y=sqrt(4+0) -> y=2#. Therefore, our point is #(0,2)#
2.) To find the derivative, we must use the Chain Rule. Start by taking the derivative of the outside function: #.5(4+sin(x))^-0.5#. Then multiply that by the derivative of the inside function, which equals #cos(x)# to get #y'=cos(x)/(2sqrt(4+sin(x)))#.
3.) Plug in #x=0# to the derivative. #y'(0)=cos(0)/(2sqrt(4+sin(0)))#. This simplifies to #y'(0)=1/(2sqrt(4+0))#, which yields #y'(0)=1/4#
4.)We use our point #(0,2)# and our calculated slope of #m=1/4# to plug into point slope form #(y-y_1)=m(x-x_1)#. We get #y_1 and x_1# from our point, so #y_1=2 and x_1=0#. This gives us #(y-2)=1/4(x-0)#. We can put this into standard form of #y=1/4x+2#
5.)We will plug #x=.12# into this equation. To make it easier, we will make #0.12# into the fraction #3/25#. So #y(.12)=(1/4)(3/25)+2#. That gives #y(.12)=(3/100)+2 -> y(.12)=2.03#.
When you use a calculator and plug #x=.12# into the original equation, you see the actual value ends up being #2.029707# which is very close to #2.03#
