Question

Please help, this chapter was on derivatives...

Answer 1

(3) Differentiating both sides of

$2x^{3/2} + y^{3/2} = 29$

with respect to x gives

$3x^{1/2} + \dfrac32 y^{1/2} \dfrac{\mathrm dy}{\mathrm dx} = 0$

Solve for dy/dx :

$\dfrac32 y^{1/2} \dfrac{\mathrm dy}{\mathrm dx} = -3x^{1/2} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{-3x^{1/2}}{\frac32y^{1/2}} = \dfrac{-2x^{1/2}}{y^{1/2}} = -2\sqrt{\dfrac xy}$

Then the slope of the tangent line to the curve at (1, 9) is

$\dfrac{\mathrm dy}{\mathrm dx} = -2\sqrt{\dfrac19} = -\dfrac23$

The equation of the tangent line would then be

y - 9 = -2/3 (x - 1)   ==>   y = -2/3 x + 29/3

(4) The slope of the tangent line to

$y=\dfrac{ax+1}{x-2}$

at a point (x, y) on the curve is

$\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{a(x-2)-(ax+1)}{(x-2)^2} = -\dfrac{2a+1}{(x-2)^2}$

When x = -1, we have a slope of 2/3, so

-(2a + 1)/(-1 - 2)² = 2/3

Solve for a :

-(2a + 1)/9 = 2/3

2a + 1 = -18/3 = -6

2a = -7

a = -7/2