Calculus Differentiation Quiz

Derivatives

• The derivative of ( f(x) = 3x^4 - 5x^3 + 2x - 7 ) is calculated using power rule:

  • ( f'(x) = 12x^3 - 15x^2 + 2 )

• For ( f(x) = \ln(x^2 + 1) ), the derivative using chain rule is:

  • ( f'(x) = \frac{2x}{x^2 + 1} )

• The second derivative of ( f(x) = e^{2x} \sin(x) ) requires the product and chain rules. First derivative:

  • ( f'(x) = e^{2x} \sin(x) \cdot 2 + e^{2x} \cos(x) )
  • Second derivative involves differentiation of the first result.

Implicit Differentiation

• For the equation ( x^2 + y^2 = 25 ), implicit differentiation gives:
  • ( 2x + 2y \frac{dy}{dx} = 0 )
  • Solving for ( \frac{dy}{dx} ) leads to ( \frac{dy}{dx} = -\frac{x}{y} )

Critical Points and Classification

• To find critical points of ( f(x) = x^3 - 6x^2 + 9x + 1 ):
  • Determine ( f'(x) ) and set it to zero to find candidates.
  • Classify each critical point using the second derivative test.

Applications of Differentiation

• The position function of a particle ( s(t) = t^3 - 6t^2 + 9t ) allows calculation of:

  • Velocity: ( v(t) = s'(t) )
  • Acceleration: ( a(t) = s''(t) )
  • Specifically at ( t = 2 ).

• The tangent line to the curve ( y = x^2 e^x ) at ( x = 0 ) involves:

  • Finding ( y(0) ) and ( f'(0) ) to establish point and slope for the line.

• The Mean Value Theorem (MVT) states that in the interval ([0, 3]) for ( f(x) = x^3 - 3x^2 ):

  • The average rate of change over this interval equals ( \frac{f(3) - f(0)}{3 - 0} ) and guarantees at least one point ( c ) in the interval where ( f'(c) ) matches this average slope.