Quotient Rule Practice Problems
Problem: Quotient Rule #1
Find the derivative of:
[latex]\Large{f(x) = \frac{4x^2}{2x-3}}[/latex]
Step 1:
We notice that the given function is a quotient of two other functions. To find the derivative of this, we can use the quotient rule. First, let’s label the two functions as:
[latex]\large{h(x) = 4x^2}[/latex] and [latex]\large{g(x) = 2x-3}[/latex]
To use the quotient rule, we also need the derivative of these functions. We can easily find the derivatives as:
[latex]\large{h'(x) = 8x}[/latex] and [latex]\large{g'(x) = 2}[/latex]
Step 2:
The quotient rule formula states that:
[latex]\large{\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{g(x)f'(x) – f(x)g'(x)}{g(x)^2}}[/latex]
Thus, substituting into the quotient rule formula, we have:
[latex]\large{\frac{d}{dx}(\frac{4x^2}{2x-3}) = \frac{(2x-3)(8x) \:- \:(4x^2)(2)}{(2x-3)^2}}[/latex]
Step 3:
Simplifying, we have:
[latex]\Large{ = \frac{16x^2-24x-8x^2}{(2x-3)^2}}[/latex]
[latex]\Large{ = \frac{8x^2-24x}{(2x-3)^2}}[/latex]
Thus, this is the final answer for the derivative. So:
[latex]\Large{f'(x) = \frac{8x^2-24x}{(2x-3)^2}}[/latex]
Solution Complete!
Problem: Quotient Rule #2
Find the derivative of:
[latex]\Large{y = \frac{x^2-3x+1}{2x^2+5}}[/latex]
Problem: Quotient Rule #3
Find the derivative of:
[latex]h(x) =\Large{ \frac{e^{x}-x}{\sin(x)}}[/latex]
Problem: Quotient Rule #4
Find the derivative of:
[latex]y = \Large{\frac{\sqrt{x}}{2x+1}}[/latex]
Problem: Quotient Rule #5
Find the derivative of:
[latex]\Large{f(x) = \frac{\ln(x)-1}{\cot(x)}}[/latex]
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