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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

$\frac {(6t-5)^5(30t+329)}{(t+9)^2}$

  • 1
    Looks like you did the top part in the wrong order for one thing – 2017-02-08
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    Using the quotient rule, the derivative, before simplifying, is given by $$\frac{36(6t-5)^5(t+9) - (1)(6t- 5)^6}{(t+9)^2}$$ The rest is simplification (e.g., factoring out $(6t-5)^5$, etc.) – 2017-02-08
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    The derivative of any quotient of functions, like the one given, can be found as follows $$\dfrac{(\text{numerator})'\,\cdot\, \text{denominator} - \text{numerator}\;\cdot\;(\text{denominator})'}{(\text{denominator})^2}$$ – 2017-02-08

5 Answers 5

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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 $$f'(x)=\frac{30t+329}{(6t-5)(t+9)}\cdot f(x)=\frac{(6t-5)^5(30t+329)}{(t+9)^2}.$$

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\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*}

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Can also use product and chain rules: $$6(6t-5)^5\cdot 6\cdot (t+9)^{-1}+(6t-5)^6\cdot -1\cdot (t+9)^{-2}\cdot 1$$

$$\frac{(t+9)\cdot 36\cdot (6t-5)^5}{t+9} -\frac{(6t-5)^6}{(t+9)^2}$$

$$\frac{(6t-5)^5[(36t+324)-(6t-5)]}{(t+9)^2}$$

$$\frac{(6t-5)^5(30t+329)}{(t+9)^2}$$

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Hint

$$\left(\frac{f(t)}{g(t)}\right)'=\frac{f'(t)g(t)-f(t)g'(t)}{(g(t))^2}$$

Take $$f(t)=(6t-5)^6 \to f'(t)=6\cdot 6\cdot(6t-5)^5\\ g(t)=t+9\to g'(t)=1$$

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    Does the order matter on the numerator? I used v'(x)u(x) - u'(x)v(x) – 2017-02-08
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    In fact, it matters! – 2017-02-08
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    @FionaLu, If you change the signal you will get the right answer! Is it clear? – 2017-02-08
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$${ \left[ \frac { (6t-5)^{ 6 } }{ t+9 } \right] }^{ \prime }=\frac { 36{ \left( 6t-5 \right) }^{ 5 }\left( t+9 \right) -(6t-5)^{ 6 } }{ { \left( t+9 \right) }^{ 2 } } =\frac { { \left( 6t-5 \right) }^{ 5 }\left[ 36\left( t+9 \right) -\left( 6t-5 \right) \right] }{ { \left( t+9 \right) }^{ 2 } } =\frac { { \left( 6t-5 \right) }^{ 5 }\left[ 30t+329 \right] }{ { \left( t+9 \right) }^{ 2 } } $$