Derivitives Help!

**mathamatics112** · February 15th 2009, 05:57 PM

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).

**Shyam** · February 15th 2009, 06:14 PM

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.

$ f(x) = \sqrt{x+1} $

slope of tangent = f'(x)

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

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

so,

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

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

$ \Rightarrow x= 8 $

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

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

eqn of tangent

y = mx + b

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

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

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

so, eqn of tangent is

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

**Shyam** · February 15th 2009, 06:22 PM

Originally Posted by mathamatics112

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

$\frac{1}{x}+\frac{1}{y}=1$

differentiating wrt x,

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

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

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

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

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