
If \[\varphi \left( x \right)\] is the inverse of the function f(x) and \[f^{'}\left( x \right) = \dfrac{1}{{\left( {1 + {x^5}} \right)}}\] then what is the value of \[\dfrac{d}{{dx}}\varphi \left( x \right)\]?
A.\[\dfrac{1}{{\left( {1 + {{\left( {\varphi \left( 5 \right)} \right)}^5}} \right)}}\]
B.\[1 + f\left( x \right)\]
C.\[\left( {1 + {{\left( {\varphi \left( 5 \right)} \right)}^5}} \right)\]
D.\[\dfrac{1}{{\left( {1 + {{\left( {f\left( 5 \right)} \right)}^5}} \right)}}\]
Answer
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Hint: We know that composite of a function and inverse return \[x\]. Apply this condition to establish a relation between \[f\left( x \right)\] and \[\phi \left( x \right)\]. Then we will derivative \[fo\varphi \left( x \right) = x\] with respect to \[x\] using chain rule. After that divide both sides by \[f'\left( {\phi \left( x \right)} \right)\]. Then find the value of \[f'\left( {\phi \left( x \right)} \right)\] by putting \[x = \phi \left( x \right)\] in \[f^{'}\left( x \right) = \dfrac{1}{{\left( {1 + {x^5}} \right)}}\]. Then we will get desire result.
Formula Used:
We will use the composite function i.e., \[f\left( {g\left( x \right)} \right)\], and derivative formula \[\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f^{'}\left( {g\left( x \right)} \right) \times g^{'}x\] and \[\dfrac{d}{{dx}}x = 1\] and by using chain rule.
Complete step by step solution:
Given \[f^{'}\left( x \right) = \dfrac{1}{{\left( {1 + {x^5}} \right)}}\], and \[\varphi \left( x \right)\] is the inverse of the function f(x),
\[ \Rightarrow f\left( x \right) = \varphi^{\prime} \left( x \right)\],
Now we will use the composite function rule, we will get,
\[ \Rightarrow \varphi \left( x \right) = x\]
Now we will differentiate on both sides we get,
\[ \Rightarrow \dfrac{d}{{dx}}f o \varphi \left( x \right) = \dfrac{d}{{dx}}x\]
\[ \Rightarrow f'\left( \varphi(x) \right) . \varphi^{\prime} (x) = 1\]
\[ \Rightarrow \dfrac{1}{{\left( {1 + {x^5}} \right)}} . \varphi^{\prime} (x) = 1\]
\[ \Rightarrow \varphi^{\prime} (x) = {1 + {(x)^5}}\]
The correct option is C.
Note: Students often do a common thing, that is they apply product rules on \[f \circ g\left( x \right)\]. They calculate the derivative \[f \circ g\left( x \right)\] as \[f'\left( x \right)g\left( x \right) + f\left( x \right)g'\left( x \right)\]. That is an incorrect way . The correct way is \[\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\].
Formula Used:
We will use the composite function i.e., \[f\left( {g\left( x \right)} \right)\], and derivative formula \[\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f^{'}\left( {g\left( x \right)} \right) \times g^{'}x\] and \[\dfrac{d}{{dx}}x = 1\] and by using chain rule.
Complete step by step solution:
Given \[f^{'}\left( x \right) = \dfrac{1}{{\left( {1 + {x^5}} \right)}}\], and \[\varphi \left( x \right)\] is the inverse of the function f(x),
\[ \Rightarrow f\left( x \right) = \varphi^{\prime} \left( x \right)\],
Now we will use the composite function rule, we will get,
\[ \Rightarrow \varphi \left( x \right) = x\]
Now we will differentiate on both sides we get,
\[ \Rightarrow \dfrac{d}{{dx}}f o \varphi \left( x \right) = \dfrac{d}{{dx}}x\]
\[ \Rightarrow f'\left( \varphi(x) \right) . \varphi^{\prime} (x) = 1\]
\[ \Rightarrow \dfrac{1}{{\left( {1 + {x^5}} \right)}} . \varphi^{\prime} (x) = 1\]
\[ \Rightarrow \varphi^{\prime} (x) = {1 + {(x)^5}}\]
The correct option is C.
Note: Students often do a common thing, that is they apply product rules on \[f \circ g\left( x \right)\]. They calculate the derivative \[f \circ g\left( x \right)\] as \[f'\left( x \right)g\left( x \right) + f\left( x \right)g'\left( x \right)\]. That is an incorrect way . The correct way is \[\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\].
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