Derivatives product rule

Front

What is the product rule in calculus?

Back

The product rule is a formula used to find the derivative of the product of two functions. If \( f(x) \) and \( g(x) \) are two differentiable functions, then the derivative of their product is given by: \( (f \cdot g)' = f' \cdot g + f \cdot g' \).

Front

If \( f(x) = u(x) \cdot v(x) \), what are \( u' \) and \( v' \)?

Back

\( u' \) and \( v' \) are the derivatives of the functions \( u(x) \) and \( v(x) \) respectively, which are used in the product rule to find the derivative of \( f(x) \).

Front

Apply the product rule to find the derivative of \( f(x) = x^2 \cdot \, \sin(x) \).

Back

Using the product rule: \( f'(x) = 2x \cdot \sin(x) + x^2 \cdot \cos(x) \).

Front

What is the derivative of \( f(x) = e^x \cdot x^3 \)?

Back

Using the product rule: \( f'(x) = e^x \cdot 3x^2 + e^x \cdot x^3 = e^x(3x^2 + x^3) \).

Front

How do you apply the product rule to find the derivative of \( f(x) = (3x^2)(\, \ln(x)) \)?

Back

Using the product rule: \( f'(x) = 6x \cdot \ln(x) + 3x^2 \cdot \frac{1}{x} = 6x \cdot \ln(x) + 3x \).

Front

What is the significance of the product rule in calculus?

Back

The product rule is significant because it allows for the differentiation of products of functions, which is essential in solving complex calculus problems involving multiple functions.

Front

Find the derivative of \( f(x) = (x^2 + 1)(x^3 - 2) \).

Back

Using the product rule: \( f'(x) = (2x)(x^3 - 2) + (x^2 + 1)(3x^2) = 2x^4 - 4x + 3x^4 + 3x^2 = 5x^4 + 3x^2 - 4x \).

Front

What is the formula for the product rule?

Back

The formula for the product rule is: \( (uv)' = u'v + uv' \), where \( u \) and \( v \) are functions of \( x \).

Front

If \( f(x) = x^4 \cdot \, an(x) \), what is \( f'(x) \)?

Back

Using the product rule: \( f'(x) = 4x^3 \cdot \tan(x) + x^4 \cdot \sec^2(x) \).

Front

What is the derivative of \( f(x) = (x^2)(e^x) \)?

Back

Using the product rule: \( f'(x) = 2x \cdot e^x + x^2 \cdot e^x = e^x(2x + x^2) \).