Find the equation of a line tngent to F(x)=sq root of (2x+1) @ the point (4,3),Explain

Question

zehanz · Answer

The slope of the tangent line at a point of a graph is equal to the derivative of the function. Here we have the function$$f(x)=\sqrt{2x+1}$$so we need to differentiate this function and then calculate f'(4).
Now f is a composition of more than one function, so we need to use the Chain Rule. 
The derivative of $$\sqrt{x} $$ is $$\frac{ 1 }{ 2\sqrt{x} }$$ therefore
$$f'(x)=\frac{ 1 }{ 2\sqrt{2x+1} }*2$$ (the *2 is the derivative of 2x+1).
This can be written as$$f'(x)=\frac{ 1 }{ \sqrt{2x+1} }$$Now calculate the derivative in x = 4 to find the slope of the tangent line:$$f'(4)=\frac{ 1 }{ \sqrt{2*4+1} }=\frac{ 1 }{ 3 }$$
The equation of the tangent line is $$y=\frac{ 1 }{3 }x + b$$We now only need to find b. Because the point (4,3) is on this line, substituting 4 for  x and 3 for y gives:$$b=y-\frac{ 1 }{ 3}x=3-\frac{ 1 }{ 3 }*4=\frac{ 8 }{ 3 }$$The equation is $$y=\frac{ 1 }{ 3 }x + \frac{ 8 }{ 3 }$$If you want to, this can also be written (by multiplying with 3 and arranging terms) as :$$x-3y+24=0$$

Hope this helps!

ZeHanz