Derivative practice problems

1. 1
   Differentiate each power

   Multiply by the exponent and reduce it by one.

   $$(4x^5)'=20x^4,\qquad(-3x^2)'=-6x$$
2. 2
   Combine the terms

   The derivative of 7 is zero.

   $$f'(x)=20x^4-6x$$

Independent check: A centered difference quotient agrees with the formula at x = 1 and x = 2.

1. 1
   Apply the product rule

   Use the derivative of each factor once.

   $$g'(x)=3x^2e^x+x^3e^x$$
2. 2
   Factor the result

   The common exponential makes the structure easy to check.

   $$g'(x)=e^x(x^3+3x^2)$$

Independent check: Expanding the factored answer and applying the product rule returns the same two terms.

1. 1
   Use the quotient rule

   Differentiate the numerator and denominator separately.

   $$h'(x)=\frac{2x(x-1)-(x^2+1)}{(x-1)^2}$$
2. 2
   Simplify the numerator

   Combine like terms without cancelling the original restriction.

   $$h'(x)=\frac{x^2-2x-1}{(x-1)^2},\quad x\ne1$$

Independent check: Rewriting h(x) as x + 1 + 2/(x - 1) gives the same derivative.

1. 1
   Differentiate the outside

   Keep the inner expression unchanged at first.

   $$5(3x^2-2)^4$$
2. 2
   Multiply by the inside derivative

   The derivative of 3x squared minus 2 is 6x.

   $$y'=5(3x^2-2)^4(6x)=30x(3x^2-2)^4$$

Independent check: At x = 1, a centered difference quotient approaches 30, matching the formula.

1. 1
   Differentiate cosine

   The outer derivative is negative sine.

   $$\frac{d}{dx}\cos(4x)=-\sin(4x)\frac{d}{dx}(4x)$$
2. 2
   Apply the inner derivative

   The derivative of 4x is 4.

   $$p'(x)=-4\sin(4x)$$

Independent check: A centered difference at x = 0 gives 0, and at x = π/8 gives -4, as the formula predicts.

1. 1
   Name the inner function

   Let u equal x squared plus 4.

   $$u=x^2+4,\qquad u'=2x$$
2. 2
   Use the logarithm rule

   Divide the inner derivative by the inner function.

   $$q'(x)=\frac{u'}u=\frac{2x}{x^2+4}$$

Independent check: At x = 1, a centered difference quotient approaches 0.4, matching 2/(1² + 4) = 0.4.

1. 1
   Differentiate both sides

   Treat y as a function of x.

   $$2x+y+xy'+2yy'=0$$
2. 2
   Isolate y prime

   Factor the derivative terms before dividing.

   $$(x+2y)y'=-(2x+y)$$
3. 3
   Divide with a condition

   The displayed slope requires x + 2y not equal to zero.

   $$\frac{dy}{dx}=-\frac{2x+y}{x+2y}$$

Independent check: Substitution into the differentiated equation makes its left side zero.

1. 1
   Find the first derivative

   Differentiate each polynomial term.

   $$r'(x)=4x^3-6x^2+1$$
2. 2
   Differentiate again

   The constant in r prime disappears.

   $$r''(x)=12x^2-12x$$

Independent check: At x = 2, the centered second difference [r(2 + h) - 2r(2) + r(2 - h)]/h² approaches 24, matching 12(2²) - 12(2) = 24.

1. 1
   Differentiate

   The derivative gives the slope function.

   $$f'(x)=3x^2-4$$
2. 2
   Evaluate at 1

   Insert the specified x-coordinate.

   $$f'(1)=3-4=-1$$

Independent check: Secant slopes through x = 1 ± h approach -1 as h approaches zero.

1. 1
   Check the left slope

   For negative x, absolute value x equals negative x.

   $$f'_-(0)=-1$$
2. 2
   Check the right slope

   For positive x, absolute value x equals x.

   $$f'_+(0)=1$$
3. 3
   Compare

   A derivative exists only when the one-sided derivative limits agree.

   $$f'_-(0)\ne f'_+(0)$$

Independent check: The graph has two distinct tangent directions at the origin.