Question

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 1

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$                  .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$                         .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$.

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

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

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.