Product Rule Practice Problems
Problem: Product Rule #1
Find the derivative of \Large{y = (4x^2+1)(3x-4)}
Step 1:
We notice that the given function is a product of two other functions. To find the derivative of this, we can use the product rule. First, let’s label the two functions as:
\large{f(x) = 4x^2+1} and \large{g(x) = 3x-4}
To use the product rule, we also need the derivative of these functions. We can easily find the derivatives as:
\large{f'(x) = 8x} and \large{g'(x) = 3}
Step 2:
The product rule formula states that:
\large{\frac{d}{dx}f(x)g(x) = f(x)g'(x) + f'(x)g(x)}
Thus, substituting into the product rule formula, we have:
\large{\frac{d}{dx}[(4x^2+1)(3x-4)] = (4x^2+1)(3) + (8x)(3x-4)}
Step 3:
Simplifying, we have:
\large{ = 12x^2+3+24x^2-32x = 36x^2-32x+3}
Thus, this is the final answer for the derivative. So:
\large{\frac{dy}{dx} = 36x^2-32x+3}
Solution Complete!
Step 1:
We notice that the given function is a product of two other functions. To find the derivative of this, we can use the product rule. First, let’s label the two functions as:
{f(x) = 4x^2+1} and {g(x) = 3x-4}
To use the product rule, we also need the derivative of these functions. We can easily find the derivatives as:
\large{f'(x) = 8x} and \large{g'(x) = 3}
Step 2:
The product rule formula states that:
{\frac{d}{dx}f(x)g(x) = f(x)g'(x) + f'(x)g(x)}
Thus, substituting into the product rule formula, we have:
\large{\frac{d}{dx}[(4x^2+1)(3x-4)]}
\large{ = (4x^2+1)(3) + (8x)(3x-4)}
Step 3:
Simplifying, we have:
\small{ = 12x^2+3+24x^2-32x = 36x^2-32x+3}
Thus, this is the final answer for the derivative. So:
\large{\frac{dy}{dx} = 36x^2-32x+3}
Solution Complete!
Problem: Product Rule #2
Find the derivative of \Large{h(x) = x\tan(x)}
Problem: Product Rule #3
Find the derivative of \Large{g(x) = (e^{3x})(x^2)}
Problem: Product Rule #4
Find the derivative of \Large{f(x) = (e^x-x)(ln(x)+1)}
Problem: Product Rule #5
Find the derivative of \Large{f(x) = sec(x)x^2}
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