Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Answer:(a) [tex]f'(x)=-\frac{2}{x^3}[/tex](b) [tex]y=-0.25x+0.75[/tex]Step-by-step explanation:The given function is [tex]f(x)=\frac{1}{x^2}[/tex]                  .... (1)According to the first principle of the derivative,[tex]f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex][tex]f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}[/tex][tex]f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}[/tex][tex]f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}[/tex][tex]f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}[/tex][tex]f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}[/tex]Cancel out common factors.[tex]f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}[/tex]By applying limit, we get[tex]f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}[/tex][tex]f'(x)=\frac{-2x)}{x^4}[/tex][tex]f'(x)=\frac{-2)}{x^3}[/tex]                         .... (2)Therefore [tex]f'(x)=-\frac{2}{x^3}[/tex].(b)Put x=2, to find the y-coordinate of point of tangency.[tex]f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25[/tex]The coordinates of point of tangency are (2,0.25).The slope of tangent at x=2 is[tex]m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}[/tex]Substitute x=2 in equation 2.[tex]f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25[/tex]The slope of the tangent line at x=2 is -0.25.The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).Using point slope form the equation of tangent is[tex]y-y_1=m(x-x_1)[/tex][tex]y-0.25=-0.25(x-2)[/tex][tex]y-0.25=-0.25x+0.5[/tex][tex]y=-0.25x+0.5+0.25[/tex][tex]y=-0.25x+0.75[/tex]Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.