05 Differentiation Techniques and Related Rates Free Online FlashCards

What does a derivative measure?

The derivative f′(x)=dy/dx is the instantaneous rate of change of f. Geometrically, f′(a) is the slope of the tangent line at x=a.

What are the fourth and fifth derivatives of x⁵−3x³+2x?

For f(x)=x⁵−3x³+2x, the fourth derivative is f⁽⁴⁾(x)=120x and the fifth derivative is f⁽⁵⁾(x)=120.

State the product rule.

(fg)′=f′g+fg′. Differentiate the first factor while leaving the second unchanged, then add the first factor times the derivative of the second.

State the quotient rule.

For g(x)≠0, (f/g)′=[f′(x)g(x)−f(x)g′(x)]/[g(x)]². The denominator is the square of the original denominator.

How does the chain rule work?

The chain rule is d/dx[f(g(x))]=f′(g(x))g′(x). Differentiate the outer function first, then multiply by the derivative of the inner function.

Differentiate y=(3x²−5)⁴.

For y=(3x²−5)⁴, y′=24x(3x²−5)³. The factor 6x comes from differentiating the inner expression 3x²−5.

Find dy/dx for x²+y²=25.

For x²+y²=25, implicit differentiation gives 2x+2y(dy/dx)=0, so dy/dx=−x/y.

What is the procedure for implicit differentiation?

Differentiate both sides, apply product and chain rules as needed, collect all dy/dx terms, factor out dy/dx, and solve for it.

State the inverse-function derivative formula.

For a one-to-one differentiable function, (f⁻¹)′(x)=1/f′(f⁻¹(x)), provided the denominator is nonzero.

Given f(2)=5 and f′(2)=3, find (f⁻¹)′(5).

If f(2)=5 and f′(2)=3, then (f⁻¹)′(5)=1/3 because f⁻¹(5)=2 and inverse slopes are reciprocals.

What is d/dx(arcsin x)?

The derivative of arcsin x is 1/√(1−x²), for −1<x<1. This follows by differentiating sin y=x implicitly.

How are velocity and acceleration related to position?

For position s(t), velocity is v(t)=s′(t), and acceleration is a(t)=s″(t). Thus acceleration is the second derivative of position.