
If $f'(4)=5,g'(4)=12,f(4)g(4)=2$ and $g(4)=6$, then find the derivative of the function $\left( \dfrac{f}{g} \right)(x)$ at the point x =4.
A. $\dfrac{5}{36}$
B. $\dfrac{11}{18}$
C. $\dfrac{23}{36}$
D. $\dfrac{13}{18}$
E. $\dfrac{19}{36}$
Answer
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Hint: In this question, you can use the quotient rule of the differentiation. The quotient rule says that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
Complete step by step answer:
Given that,
The derivative of the function $f(x)$ at the point x = 4 is 5
The derivative of the function $g(x)$ at the point x = 4 is 12
The product of the function $f(x)g(x)$ at the point x= 4 is 2
The value of the function $g(x)$ at the point x = 4 is 6
Symbolically, $f'(4)=5,g'(4)=12,f(4)g(4)=2$ and $g(4)=6$
To find the derivative of the function $\left( \dfrac{f(x)}{g(x)} \right)$ at the point x = 4 by using quotient rule of differentiation. The quotient rule says that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
Symbolically, $\dfrac{d}{dx}\left( \dfrac{f(x)}{g(x)} \right)=\dfrac{g(x)f'(x)-f(x)g'(x)}{{{\left( g(x) \right)}^{2}}}........(1)$
Now put x = 4 in the equation (1), we get
$\dfrac{d}{dx}\left( \dfrac{f}{g} \right)(x)=\dfrac{g(4)f'(4)-f(4)g'(4)}{{{\left( g(4) \right)}^{2}}}........(1)$
For the value of the function $f(x)$ at the point x =4,
$f(4)g(4)=2\Rightarrow f(4)=\dfrac{2}{g(4)}$
Put the value of the $g(4)=6$ , we get
$f(4)=\dfrac{2}{6}=\dfrac{1}{3}$
Now put all the values in the question (1), we get
${{\left( \dfrac{f}{g} \right)}^{'}}(4)=\dfrac{6\times 5-\left( \dfrac{1}{3} \right)\times 12}{{{\left( 6 \right)}^{2}}}=\dfrac{30-4}{36}=\dfrac{26}{36}=\dfrac{13}{18}$
Hence, the derivative of the function $\left( \dfrac{f}{g} \right)(x)$ at the point x =4 is $\dfrac{13}{18}$.
Therefore, the correct option for the given question is option (D).
Note: Students should remember the formula of quotient rule of differentiation i.e $\dfrac{d}{dx}\left( \dfrac{f(x)}{g(x)} \right)=\dfrac{g(x)f'(x)-f(x)g'(x)}{{{\left( g(x) \right)}^{2}}}$ for solving these types of problems.Most of the students get confuse to find value of $f(4)$ So, we have to take from the given data i.e $f(4) g(4)=2$.
Complete step by step answer:
Given that,
The derivative of the function $f(x)$ at the point x = 4 is 5
The derivative of the function $g(x)$ at the point x = 4 is 12
The product of the function $f(x)g(x)$ at the point x= 4 is 2
The value of the function $g(x)$ at the point x = 4 is 6
Symbolically, $f'(4)=5,g'(4)=12,f(4)g(4)=2$ and $g(4)=6$
To find the derivative of the function $\left( \dfrac{f(x)}{g(x)} \right)$ at the point x = 4 by using quotient rule of differentiation. The quotient rule says that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
Symbolically, $\dfrac{d}{dx}\left( \dfrac{f(x)}{g(x)} \right)=\dfrac{g(x)f'(x)-f(x)g'(x)}{{{\left( g(x) \right)}^{2}}}........(1)$
Now put x = 4 in the equation (1), we get
$\dfrac{d}{dx}\left( \dfrac{f}{g} \right)(x)=\dfrac{g(4)f'(4)-f(4)g'(4)}{{{\left( g(4) \right)}^{2}}}........(1)$
For the value of the function $f(x)$ at the point x =4,
$f(4)g(4)=2\Rightarrow f(4)=\dfrac{2}{g(4)}$
Put the value of the $g(4)=6$ , we get
$f(4)=\dfrac{2}{6}=\dfrac{1}{3}$
Now put all the values in the question (1), we get
${{\left( \dfrac{f}{g} \right)}^{'}}(4)=\dfrac{6\times 5-\left( \dfrac{1}{3} \right)\times 12}{{{\left( 6 \right)}^{2}}}=\dfrac{30-4}{36}=\dfrac{26}{36}=\dfrac{13}{18}$
Hence, the derivative of the function $\left( \dfrac{f}{g} \right)(x)$ at the point x =4 is $\dfrac{13}{18}$.
Therefore, the correct option for the given question is option (D).
Note: Students should remember the formula of quotient rule of differentiation i.e $\dfrac{d}{dx}\left( \dfrac{f(x)}{g(x)} \right)=\dfrac{g(x)f'(x)-f(x)g'(x)}{{{\left( g(x) \right)}^{2}}}$ for solving these types of problems.Most of the students get confuse to find value of $f(4)$ So, we have to take from the given data i.e $f(4) g(4)=2$.
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