find equation of tangent line of the function f(x)= (x-2)^1/2 at point (18,4)

Question

Find the tangent line equation in form y=mx+b

Answer 1

f(x) = (x-2)^1/2

We know that d/dx(x^n) = nx^n-1

Apply derivative to each sides with respect of x.

d/dx f (x) = d/dx(x-2)^1/2

f '(x) = 1/2(x-2)^1/2-1

= 1/2(x-2)^-1/2

= 1/2(1/(x-2)^1/2)

= (1/2)/sqrt(x-2)

f '(18) = (1/2)/sqrt(18-2)

=(1/2)/sqrt16

= (1/2)/4

= 1/8

Tangent line slope is m and point(18,4).

Here f '(18) = m

Tangent line equation is y-4 = 1/8(x-18)

8(y-4) = x-18

8y-32 = x-18

Add 32 to each side.

8y-32+32 = x-18+32

8y  = x+14

Divide to each side by 8.

8y/8 = (x+14)/8

y = x/8+14/8

y = x/8+7/4

Required line equation is y = x/8+7/4.