Let f be a function defined for all x - R If f is differentiable and f x3 -x5 for all x - R x- 0 then the value of f-27 is

The correct option is A 15 Given, f ( x^3 )=x^5 Differentiating w.r.t x we get, f' ( x^3 )×3x^2=5x^4⇒ f'(x^3)=53x^2 Putting x=3 f ...

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Standard XII

Mathematics

Chain Rule of Differentiation

Let f be a ...

Question

Let $f$ be a function defined for all $x ϵ R$ If f is differentiable and $f (x^{3}) = x^{5}$ for all $x ϵ R (x \neq 0)$ then the value of $f^{'} (27)$ is

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Solution

The correct option is A $15$
Given, $f (x^{3}) = x^{5}$
Differentiating w.r.t $x$ we get,
$f^{'} (x^{3}) \times 3 x^{2} = 5 x^{4} \Rightarrow f^{'} (x^{3}) = \frac{5}{3} x^{2}$
Putting $x = 3$
$f^{'} (27) = \frac{5}{3} \times 3^{2} = 15$

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Let f be a function defined for all xϵR If f is differentiable and f(x3)=x5 for all xϵR(x≠0) then the value of f′(27) is

The correct option is A 15Given, f(x3)=x5Differentiating w.r.t x we get,f′(x3)×3x2=5x4⇒f′(x3)=53x2Putting x=3f′(27)=53×32=15

Let $f$ be a function defined for all $x ϵ R$ If f is differentiable and $f (x^{3}) = x^{5}$ for all $x ϵ R (x \neq 0)$ then the value of $f^{'} (27)$ is

The correct option is A $15$
Given, $f (x^{3}) = x^{5}$
Differentiating w.r.t $x$ we get,
$f^{'} (x^{3}) \times 3 x^{2} = 5 x^{4} \Rightarrow f^{'} (x^{3}) = \frac{5}{3} x^{2}$
Putting $x = 3$
$f^{'} (27) = \frac{5}{3} \times 3^{2} = 15$