Calculus Differentiation Quiz

Study Notes

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.

Description

Test your skills on various differentiation techniques including finding derivatives, using implicit differentiation, and analyzing critical points. This quiz covers essential concepts such as the first and second derivatives and their applications in real-world scenarios. Perfect for students aiming to strengthen their understanding of calculus.