Let f(x + y) = f(x) f(y) and f(x) = 1 + sin (3x) g(x), where g is differentiable. The f ′(x) is equal to

Question

Let f(x + y) = f(x) f(y) and f(x) = 1 + sin (3x) g(x), where g is differentiable. The f ′(x) is equal to (A) 3f(x) (B) g(0) (C) f(x)g(0) (D) 3g(x)

Tardigrade Admin · Accepted Answer

f'(x) = displaystyle lim h arrow 0 (f(x +h)-f(x)/h) = displaystyle lim h arrow 0 (f(x) f(h)-f(x)/h) =f(x) displaystyle lim h arrow 0((1+ sin 3 h(g(h))-1)/h)) =f(x) displaystyle lim h arrow 0 ( sin 3 h/3 h) displaystyle lim h arrow 0 g(h) =f(x) × 1 × g(0)=f(x) g(0)

Q. Let $f (x + y) = f (x) f (y)$ and $f (x) = 1 + sin (3 x) g (x)$ , where $g$ is differentiable. The $f' (x)$ is equal to

$3 f (x)$

$g (0)$

$f (x) g (0)$

$3 g (x)$

$3 f (x) g (0)$

$f^{^{'}} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$
$= h \to 0 lim \frac{f ( x ) f ( h ) - f ( x )}{h}$
$= f (x) h \to 0 lim (\frac{1 + sin 3 h ( g ( h )) - 1 )}{h})$
$= f (x) h \to 0 lim \frac{sin 3 h}{3 h} h \to 0 lim g (h)$
$= f (x) \times 1 \times g (0) = f (x) g (0)$

Q. Let f(x+y)=f(x)f(y) and f(x)=1+sin(3x)g(x), where g is differentiable. The f′(x) is equal to

3f(x)

g(0)

f(x)g(0)

3g(x)

3f(x)g(0)

f′(x)=h→0lim​hf(x+h)−f(x)​ =h→0lim​hf(x)f(h)−f(x)​ =f(x)h→0lim​(h1+sin3h(g(h))−1)​) =f(x)h→0lim​3hsin3h​h→0lim​g(h) =f(x)×1×g(0)=f(x)g(0)