# Thread: tricky derivatives

1. ## tricky derivatives

1. Let f(x)= -x^3 - 2x^2 + x + 1 and g(x)= ln(x+1)+1
a. Find the equation of the line tangent to f at x=0
b. Show that g has the same tangent line as f at x=0

2. Given x=t^4 - t^2 + t and y=2t^3 - t, find dy/dx

3. Given x^2 + 3y^2 = 1
a. Find y'
b. Find the slope of the curve at the point (.5,.5)
c. Find the equation of the line tangent to the ellipse at (.5,.5)

2. Originally Posted by Dr. Noobles
2. Given x=t^4 - t^2 + t and y=2t^3 - t, find dy/dx
You find the parameter derivative,
$\frac{dy}{dt}=6t^2-1$

And you find the parameter derivative,
$\frac{dx}{dt}=4t^3-2t+1$

Thus,
$\frac{dy}{dx}=\frac{6t^2-1}{4t^3-2t+1}$

3. Originally Posted by Dr. Noobles
1. Let f(x)= -x^3 - 2x^2 + x + 1 and g(x)= ln(x+1)+1
a. Find the equation of the line tangent to f at x=0
b. Show that g has the same tangent line as f at x=0
a) $f'(x) = -3x^2 - 4x + 1$

So the slope of the tangent line at x = 0 is:
$m = f'(0) = 1$

The tangent line is thus: $y = x + b$ where b is the intercept. This line touches the function y = f(x), and we know that y = f(0) = 1. So the line $y = x + b$ passes through the point (0, 1). Thus
$1 = 0 + b$ gives b = 1.

Thus your tangent line is $y = x + 1$. (See graph below.)

b) To show that g(x) has the same tangent line at x = 0, all we need to do is verify that g(0) = 1 and that g'(0) = 1.

So:
$g(0) = ln(x+1) + 1 = ln(0 + 1) + 1 = ln(1) + 1 = 0 + 1 = 1$ (Check!)
$g'(x) = \frac{1}{x+1}$
so
$g'(x) = \frac{1}{0 + 1} = 1$ (Check!)

-Dan

4. Originally Posted by Dr. Noobles
3. Given x^2 + 3y^2 = 1
a. Find y'
b. Find the slope of the curve at the point (.5,.5)
c. Find the equation of the line tangent to the ellipse at (.5,.5)
a) Use implicit differentiation:
$x^2 + 3y^2 = 1$

$2x + 6yy' = 0$

$6yy' = -2x$

$y' = -\frac{x}{3y}$

b) At $\left ( \frac{1}{2}, \frac{1}{2} \right )$:

$y' = -\frac{\frac{1}{2}}{3\frac{1}{2}} = -\frac{1}{3}$

c) We know the slope is $m = -\frac{1}{3}$ and the tangent line passes through the point $\left ( \frac{1}{2}, \frac{1}{2} \right )$ so:
$y = mx + b$

$\frac{1}{2} = -\frac{1}{3} \cdot \frac{1}{2} + b$

$b = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$

So the tangent line is
$y = -\frac{1}{3}x + \frac{2}{3}$

-Dan