022 Exam 2 Sample B, Problem 1

Find the derivative of  $y\,=\,\ln {\frac {(x+1)^{4}}{(2x-5)(x+4)}}.$

Foundations:
This problem requires several advanced rules of differentiation. In particular, you need
The Chain Rule: If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are differentiable functions, then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (f\circ g)'(x) = f'(g(x))\cdot g'(x).}
The Product Rule: If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are differentiable functions, then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).}
The Quotient Rule: If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are differentiable functions and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x) \neq 0}  , then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. }
Additionally, we will need our power rule for differentiation:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(x^n\right)'\,=\,nx^{n-1},} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n\neq 0} ,
as well as the derivative of natural log:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\ln x\right)'\,=\,\frac{1}{x}.}

 Solution:

Step 1:
We need to identify the composed functions in order to apply the chain rule. Note that if we set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)\,=\,\ln x} , and
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)\,=\,\frac{(x+1)^4}{(2x - 5)(x + 4)},}
we then have  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).}

Step 2:
We can now apply all three advanced techniques. For Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)} , we can use both the quotient and product rule to find
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} f'(x)&=&\displaystyle{\frac{((x+1)^4)'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^4}{(2x-5)^2(x+4)^2}} \\ \\ &=&\displaystyle{\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}.} \end{array}}

Step 3:
We can now use the chain rule to find
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} y' & = & \left(g\circ f\right)'(x)\\ \\ & = & g'\left(f(x)\right)\cdot f'(x)\\ \\& = &\displaystyle{\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{((x+1)^4)'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^4}{(2x-5)^2(x+4)^2}} \\ \\ &=&\displaystyle{\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}. } \end{array}} Note that many teachers do not prefer a cleaned up answer, and may request that you do not simplify. In this case, we could write the answer as
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'=\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}. }

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'=\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}. }

Return to Sample Exam