Find the indicated derivative
$\frac {d}{dt}$ $\frac {(6t-5)^6}{t+9}$
I'm stuck after getting to this part
$\frac {(6t-5)^6-36(6t-5)^5(t+9)}{(t+9)^2}$
How do they get to the answer
$\frac {(6t-5)^5(30t+329)}{(t+9)^2}$
$\endgroup$ 35 Answers
$\begingroup$\begin{align*} \frac{d}{dt}\left(\frac{(6t-5)^6}{t+9} \right) & = \frac{6((6t-5)^5.6)(t+9)-(6t-5)^6}{(t+9)^2}\\ & = \frac{(6t-5)^5(36(t+9)-6t+5)}{(t+9)^2}\\ & = \frac{(6t-5)^5(30t+329)}{(t+9)^2} \end{align*}
$\endgroup$ $\begingroup$They simply factor out $(6t-5)^5$: $$\frac {(6t-5)^6-36(6t-5)^5(t+9)}{(t+9)^2}=\frac{(6t−5)^5(6t−5-36(t+9))}{(t+9)^2}. $$ (Note there seems o be a sign error in your computation)
Faster with logarithmic differentiation: $$\frac{f'(x)}{f(x)}=\frac{36}{6t-5}-\frac1{t+9}=\frac{30t+329}{(6t-5)(t+9)}, $$ so $$f'(x)=\frac{30t+329}{(6t-5)(t+9)}\cdot f(x)=\frac{(6t-5)^5(30t+329)}{(t+9)^2}.$$
$\endgroup$ $\begingroup$Can also use product and chain rules: $$6(6t-5)^5\cdot 6\cdot (t+9)^{-1}+(6t-5)^6\cdot -1\cdot (t+9)^{-2}\cdot 1$$
$$\frac{(t+9)\cdot 36\cdot (6t-5)^5}{t+9} -\frac{(6t-5)^6}{(t+9)^2}$$
$$\frac{(6t-5)^5[(36t+324)-(6t-5)]}{(t+9)^2}$$
$$\frac{(6t-5)^5(30t+329)}{(t+9)^2}$$
$\endgroup$ $\begingroup$Hint
$$\left(\frac{f(t)}{g(t)}\right)'=\frac{f'(t)g(t)-f(t)g'(t)}{(g(t))^2}$$
Take $$f(t)=(6t-5)^6 \to f'(t)=6\cdot 6\cdot(6t-5)^5\\ g(t)=t+9\to g'(t)=1$$
$\endgroup$ 3 $\begingroup$$${ \left[ \frac { (6t-5)^{ 6 } }{ t+9 } \right] }^{ \prime }=\frac { 36{ \left( 6t-5 \right) }^{ 5 }\left( t+9 \right) -(6t-5)^{ 6 } }{ { \left( t+9 \right) }^{ 2 } } =\frac { { \left( 6t-5 \right) }^{ 5 }\left[ 36\left( t+9 \right) -\left( 6t-5 \right) \right] }{ { \left( t+9 \right) }^{ 2 } } =\frac { { \left( 6t-5 \right) }^{ 5 }\left[ 30t+329 \right] }{ { \left( t+9 \right) }^{ 2 } } $$
$\endgroup$
