If f:R→R is a function defined as fx^3=x^5, ∀x∈R-0 and fx is differentiable ∀x∈R, then the value of 1/4f^(')27 is equal to (here f^(') represents the…

Question

If f:R→R is a function defined as fx^3=x^5, ∀x∈R-0 and fx is differentiable ∀x∈R, then the value of 1/4f^(')27 is equal to (here f^(') represents the derivative of f)

MARKS Content Team · Accepted Answer

Correct answer: 3.75. fx^3=x^5 On differentiating with respect to x f^(')x^3⋅3x^2=5⋅x^4 f^(′) x^3 =5/3 x^2 Putting x=3, we get, f^(′) 27 =5/3 9 =15

If $f : R \to R$ is a function defined as $f (x^{3}) = x^{5}, \forall x \in R - \{0\}$ and $f (x)$ is differentiable $\forall x \in R$ , then the value of $\frac{1}{4} f^{'} (27)$ is equal to (here $f^{'}$ represents the derivative of $f$ )

Solution

More Differentiation Questions

If f:R→R is a function defined as fx3=x5, ∀x∈R-0 and fx is differentiable ∀x∈R, then the value of 14f'27 is equal to (here f' represents the derivative of f)

Solution

More Differentiation Questions

If $f : R \to R$ is a function defined as $f (x^{3}) = x^{5}, \forall x \in R - \{0\}$ and $f (x)$ is differentiable $\forall x \in R$ , then the value of $\frac{1}{4} f^{'} (27)$ is equal to (here $f^{'}$ represents the derivative of $f$ )