Logarithmic differentiation
Inverse functions
Inverse-trig derivatives
Implicit differentiatioon
Tangent lines and related rates
100

What should you do first when using logarithmic differentiation

Take the natural logarithm of both sides

100

What is teh relationship between the graphs of f and f-1

They are reflections across the line y=x

100

Find the derivative: f(x) = arcsin(x)

fl(x) = 1 / sqr(1-x2)

100

Differentiate implicitly: x2+y2 = 25

2x + 2y(dy/dx) = 0

So, dy/dx = -x/y

100

What is the slope of the tangent line to f(x) at x=a

m = fl(a)

200

y = xx

yl = xx(lnx+1)

200

State the derivative formula for an inverse function

(f-1)(a) = 1 / fl(f-1(a))

200

Find the derivative: f(x) = arccos(x)

fl(x) = - 1 / sqr(1-x2)

200
Differentiate: x3+y3 = 10

3x2+3y2(dy/dx) = 0

Therefore, dy/dx = - x2/y2

200

Find the tangent line to f(x) = x2 at x=2

The point is (2,4) and the slope is fl(2) = 4

Thus, y-4 = 4(x-2) or y= 4x-4

300

Differentiate: y = (x2+1)3 / (x-4)2

yl= (x2+1)3 / (x-4)2 (6x / x2 + 1  - 2 / x-4)

300

If f(2) = 5 and fl(2) = 4 find (f-1)l(5)

Since f(2) = 5 then f-1(5) = 2, therefore f-1 (5) = 1/4

300

Find the derivative: f(x) = arctan(x)

fl(x) = 1 / 1+x2

300

Find the slope of the curve: x2+y2 = 25 at the point (3,4)

Since dy/dx = -x/y

We get dy/dx = -3/4

300

Find the tangent line to f(x) sqr(4x2 + 5) at x=1

The point is (1,3)

The derivative is fl(x) = 4x / sqr(4x2+5)

So fl(1) = 4/3

Therefore, y-3 = 4/3 (x-1)

400

Differentiate: y = xsinx

yl = xsinx(cos(x)lnx + sin(x) / x)

400

Find the inverse of: f(x) = 3x-7

Let y=3x-7. Solve for x: x = y+7 / 3 thus, f-1 (x) = x+7/3

400

Differentiate: f(x) = arctan(3x)

fl(x) = 3 / 1+9x2

400
Find dy/dx: xy+y2 = 6

Differentiate: x(dy/dx) + y + 2y(dy/dx) = 0

Therefore, dy/dx = -y / x+2y

400

A circle's radius increases at 3cm/s. How fast is the circumference changing

C = 2(pi)(r)

We have dC/dt = 2(pi)(r) (dr/dt)

Thus, dC/dt = 6(pi) cm/s

500

Use logarithmic differentiation to find yl : y = sqr(x2+1) (x-2)5  / e3x(x+1)4

yl = sqr(x2+1) (x-2)5  / e3x(x+1)4  (x/x2+1  + 5/x-2  - 3 - 4/x+1)

500

If f(x) = x3 + 1 find (f-1)l(9)

First solve f(x) = 9 -> x3 + 1 = 9 -> x3 = 8 -> x=2

Then, fl(x) = 3x+2 so fl(2) = 12 -> (f-1)l(9) = 1/12

500

Differentiate: f(x) = arcsin(x2)

fl(x) = 2x / sqr(1-x4)

500
Find the tangent line to: x2+y2 = 25 at (3,4)

The slope is m = -3/4

Using point-slope form: y-4 = -3/4 (x-3)

Therefore in point-slope form: y = -3/4x + 25/4

500

A 10fit ladder leans against a wall. Its bottom moves away from the wall at 2ft/s. How fast is the top moving when the bottom is 6ft from the wall.

The ladder relationship is x2+y2 = 100

When x=6 then y =8

Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0

Substitute: 2(6)(2) + 2(8)(dy/dt) = 0

dy/dt = -3/2 ft/s