The first order derivative of f(x) is given by f'(x) = sec4x + 5 with f(0) = 0 . Then f(x) =

Question

The first order derivative of f(x) is given by f'(x) = sec4x + 5 with f(0) = 0 . Then f(x) =  (A)  tan x+(1/3)tan3 x+5x+2  (B)  tan x+(1/3)tan3 x+5x  (C)  tan x+(1/3)tan3 x  (D)  (1/3) tan3 x+5x

Tardigrade Admin · Accepted Answer

We have, f'(x) = sec4 x + 5 ⇒ ∫ f'(x) dx = ∫ sec4 x dx + ∫ 5 dx ⇒ f(x) = tan x + (1/3) tan3 x + 5x +c Now, f(0) = tan(0) + (1/3) tan3(0) + 5 × 0 + c = 0 ⇒ c = 0

Q. The first order derivative of $f (x)$ is given by $f^{'} (x) = se c^{4} x + 5$ with $f (0) = 0$ . Then $f (x) =$

$t an x + \frac{1}{3} t a n^{3} x + 5 x + 2$

$t an x + \frac{1}{3} t a n^{3} x + 5 x$

$t an x + \frac{1}{3} t a n^{3} x$

$\frac{1}{3} t a n^{3} x + 5 x$

We have, $f^{'} (x) = se c^{4} x + 5$
$\Rightarrow \int f^{'} (x) d x = \int se c^{4} x d x + \int 5 d x$
$\Rightarrow f (x) = t an x + \frac{1}{3} t a n^{3} x + 5 x + c$
Now, $f (0) = t an (0) + \frac{1}{3} t a n^{3} (0) + 5 \times 0 + c = 0$
$\Rightarrow c = 0$

Q. The first order derivative of f(x) is given by f′(x)=sec4x+5 with f(0)=0 . Then f(x)=

tanx+31​tan3x+5x+2

tanx+31​tan3x+5x

tanx+31​tan3x

31​tan3x+5x