You visited us 96 times! Enjoying our articles? Unlock Full Access!

Definition of a Determinant

If fx=e x...

Question

If $f (x) = e$ $^{x}$ $g (x),$
$g (0) = 1, g^{'} (0) = 3$ , then $f^{'} (0)$ is

0

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

4

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

3

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

7

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

Open in App

Solution

The correct option is D $4$
$f^{'} (x) = e^{^{x}} g (x) + e^{^{x}} g^{'} (x)$
For x=0:
$f^{'} (0) = g (0) + g^{'} (0) = 4$

Suggest Corrections

Similar questions

f (x) = e^{x} g (x), g (0) = 2, g^{'} (0) = 1

f^{'} (0)

f (x) = sin (x + a)

g (x) = b sin x + c cos x

f (0) = g (0)

f^{'} (0) = g^{'} (0)

L e t f (x) = \frac{g (x)}{x} w h e n x \neq 0 a n d f (0) = 0 . I f g (0) = g^{1} (0) a n d g^{11} (0) = 17 t h e n f^{1} (0)

f

f (0) = 0

g (x) = \frac{f (x)}{x}

g (0)

g

0

g^{''} (x)

g (0) \neq 0

g^{'} (0) = 0

g^{n} (0) \neq 0

f (x) = g (x) sin x .

lim x \to 0 [g (x) cot x - g (0) c o s e c x] = f^{''} (0)

f^{'} (0) = g (0)

Join BYJU'S Learning Program