1. ## Derivitives Help!

Find an equation of the straight line that is tangent to the graph of f(x) root(x+1) and parallel to x - 6y+4 = 0.

Find the slope of the tangent to the curve 1/x + 1/y = 1 at the point (2,2).

2. Originally Posted by mathamatics112
Find an equation of the straight line that is tangent to the graph of f(x) = root(x+1) and parallel to x - 6y+4 = 0.
$\displaystyle f(x) = \sqrt{x+1}$

slope of tangent = f'(x)

$\displaystyle f'(x) = \frac{-1}{2\sqrt{x+1}}$

slope of tangent, which is parallel to line x - 6y+4 = 0 , is $\displaystyle \frac{1}{6}$

so,

$\displaystyle f'(x) = \frac{-1}{2\sqrt{x+1}}=\frac{1}{6}$

$\displaystyle \sqrt{x+1}=-3$

$\displaystyle \Rightarrow x= 8$

$\displaystyle y = \sqrt{8+1}=3$

so, (8, 3) is the point on the curve where tangent is drawn.

eqn of tangent

y = mx + b

$\displaystyle y = \frac{1}{6}x + b$

$\displaystyle 3 = \frac{1}{6}(8) + b$

$\displaystyle b = \frac{5}{3}$

so, eqn of tangent is

$\displaystyle y = \frac{1}{6}x + \frac{5}{3}$

3. Originally Posted by mathamatics112
Find the slope of the tangent to the curve 1/x + 1/y = 1 at the point (2,2).
$\displaystyle \frac{1}{x}+\frac{1}{y}=1$

differentiating wrt x,

$\displaystyle \frac{-1}{x^2}-\frac{1}{y^2}y'=0$

$\displaystyle y' =\frac{-y^2}{x^2}$

$\displaystyle y' \;at\; (2, 2) =\frac{-(2)^2}{(2)^2}=-1$

so, slope of tangent at (2, 2) is -1

eqn of tangent

y = mx + b

2 = -1(2) + b

b = 4

eqn of tangent is

y = -x + 4