Let gx - x5 lnfx where fx is adifferentiable positive function on 0- satisfying f3 - 1 and f-3 - 2 then g-3 is equal to

The correct option is D 486 Given, g(x) = x^5 ln(f(x)) Differentiating w.r.t .x we get, ⇒ g'(x) = 5x^4 . ln f(x) + x^5 f'(x)/f(x) Put #160;x ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

Let gx = x5...

Question

Let $g (x) = x^{5} ln (f (x))$ where $f (x)$ is adifferentiable positive function on $(0, \infty)$ satisfying $f (3) = 1$ and $f^{'} (3) = 2$ then $g^{'} (3)$ is equal to

243

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

486

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $486$
Given, $g (x) = x^{5} ln (f (x))$
Differentiating w.r.t .x we get,
$\Rightarrow g^{'} (x) = 5 x^{4} . ln f (x) + \frac{x^{5} f^{'} (x)}{f (x)}$
Put $x = 3$
$\Rightarrow g^{'} (3) = 5. (3)^{4} ln (1) + 3^{5} .2 = 486$

Suggest Corrections

Similar questions

Q. Let

g (x) = x^{3} ln (x^{2} f (x))

, where

f (x)

is a differentiable, positive function on

(0, \infty)

satisfying

f (2) = \frac{1}{4}

and

f^{'} (2) = - 3

, then

g^{'} (2)

equal to

Q. Let

g (x)

be a polynomial of degree one and

f (x)

be defined by

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} g (x), & x \leq 0 {[\frac{1 + x}{2 + x}]}^{1 / x}, & x > 0 \end{matrix}

Let

f (x)

be a continuous function satisfying

f^{'} (1) = f (- 1) .

Then

f (- 2)

is equal to

Q. Let

f : R \to R

be a differentiable function satisfying

f^{'} (3) + f^{'} (2) = 0

. Then

lim x \to 0 {(\frac{1 + f (3 + x) - f (3)}{1 + f (2 - x) - f (2)})}^{1 / x}

is equal to :

Q.
Let

g (x) = log (f (x))

where

f (x)

is a twice differentiable positive function on

(0, \infty)

such that

f (x + 1) = x f (x)

. Then, for

N = 1, 2, 3, \dots g^{''} (N + \frac{1}{2}) - g^{''} (\frac{1}{2}) =

Q. Let

f : R \to R

be a differentiable function satisfying

f^{'} (3) + f^{'} (2) = 0

. Then

lim x \to 0 {(\frac{1 + f (3 + x) - f (3)}{1 + f (2 - x) - f (2)})}^{1 / x}

is equal to :

Join BYJU'S Learning Program

Let g(x)=x5ln(f(x)) where f(x) is adifferentiable positive function on (0,∞) satisfying f(3)=1 and f′(3)=2 then g′(3) is equal to

The correct option is D 486Given, g(x)=x5ln(f(x))Differentiating w.r.t .x we get,⇒g′(x)=5x4.lnf(x)+x5f′(x)f(x)Put x=3⇒g′(3)=5.(3)4ln(1)+35.2=486

Let $g (x) = x^{5} ln (f (x))$ where $f (x)$ is adifferentiable positive function on $(0, \infty)$ satisfying $f (3) = 1$ and $f^{'} (3) = 2$ then $g^{'} (3)$ is equal to

The correct option is D $486$
Given, $g (x) = x^{5} ln (f (x))$
Differentiating w.r.t .x we get,
$\Rightarrow g^{'} (x) = 5 x^{4} . ln f (x) + \frac{x^{5} f^{'} (x)}{f (x)}$
Put $x = 3$
$\Rightarrow g^{'} (3) = 5. (3)^{4} ln (1) + 3^{5} .2 = 486$