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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

If f: R → R is a function defined as f( x ^ 3 ) = x ^ 5 , x ∈ R - \...

If f:R→Rf: R \to R is a function defined as f(x3)=x5,∀x∈R−{0}andf\left({x}^{3}\right) = {x}^{5}, \forall x \in R - \left\{0\right\} andf\left(x\right)isdifferentiableis differentiable\forall x \in R,thenthevalueof, then the value of\dfrac{1}{4}{f}^{'}(27)\left(27\right) is equal to (here f′{f}^{'} represents the derivative of ff)

Correct answer

3.75

Step-by-step solution

f(x3)=x5f\left({x}^{3}\right) = {x}^{5}
On differentiating with respect to xx
f′(x3)⋅3x2=5⋅x4{f}^{'}\left({x}^{3}\right) ⋅ 3{x}^{2} = 5 ⋅ {x}^{4}
f′(x3)=53x2{f}^{′} \left({x}^{3}\right) = \dfrac{5}{3} {x}^{2}
Putting x=3x = 3, we get,
f′(27)=53(9)=15{f}^{′} \left(27\right) = \dfrac{5}{3} \left(9\right) = 15
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