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 $f$ and $g$ are differentiable functions, then
$(f\circ g)'(x)=f'(g(x))\cdot g'(x).$
The Product Rule: If $f$ and $g$ are differentiable functions, then
$(fg)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x).$
The Quotient Rule: If $f$ and $g$ are differentiable functions and $g(x)\neq 0$ , then
$\left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^{2}}}.$
Additionally, we will need our power rule for differentiation:
$\left(x^{n}\right)'\,=\,nx^{n-1},$ for $n\neq 0$,
as well as the derivative of natural log:
$\left(\ln x\right)'\,=\,{\frac {1}{x}}.$

 Solution:

Step 1:
We need to identify the composed functions in order to apply the chain rule. Note that if we set $g(x)\,=\,\ln x$, and
$f(x)\,=\,{\frac {(x+1)^{4}}{(2x-5)(x+4)}},$
we then have  $y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).$

Step 2:
We can now apply all three advanced techniques. For $f'(x)$, we must use both the quotient and product rule to find
${\begin{array}{rcl}f'(x)&=&{\frac {((x+1)^{4})'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^{4}}{(2x-5)^{2}(x+4)^{2}}}\\&=&{\frac {(4(x+1)^{3})(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^{4}}{(2x-5)^{2}(x+4)^{2}}}.\end{array}}$

Step 3:
We can now use the chain rule to find
${\begin{array}{rcl}y'&=&\left(g\circ f\right)'(x)\\\\&=&g'\left(f(x)\right)\cdot f'(x)\\\\&=&\left[{\frac {(2x-5)(x+4)}{(x+1)^{4}}}\right]{\frac {((x+1)^{4})'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^{4}}{(2x-5)^{2}(x+4)^{2}}}\\\\&=&\left[{\frac {(2x-5)(x+4)}{(x+1)^{4}}}\right]{\frac {(4(x+1)^{3})(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^{4}}{(2x-5)^{2}(x+4)^{2}}}\end{array}}$ 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
$y'=\left[{\frac {(2x-5)(x+4)}{(x+1)^{4}}}\right]{\frac {(4(x+1)^{3})(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^{4}}{(2x-5)^{2}(x+4)^{2}}}$

Final Answer:
$y'=\left[{\frac {(2x-5)(x+4)}{(x+1)^{4}}}\right]{\frac {(4(x+1)^{3})(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^{4}}{(2x-5)^{2}(x+4)^{2}}}$

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