Find $f ^ { t } ( x )$
$f ( x ) = 8 e ^ { x } - 3 x ^ { 2 } \arctan ( x )$
A)
$f ^ { t } ( x ) = 8 e ^ { x } - \frac { 3 x ^ { 2 } } { 1 + x ^ { 2 } } - 2 x ( \arctan ( x ) )$
B)
$f ^ {t } ( x ) = 8 e ^ { x } - \frac { 3 x ^ { 2 } } { 1 + x ^ { 2 } } + 6 x ( \arctan ( x ) )$
C)
$f ^ { t } ( x ) = 8 e ^ { x } - \frac { 3 x ^ { 2 } } { 1 + x ^ { 2 } } - 6 x ( \arcsin ( x ) )$
D)
$f ^ { t } ( x ) = 8 e ^ { x } - \frac { 3 x ^ { 2 } } { 1 + x ^ { 2 } } - 6 x ( \arctan ( x ) )$
E) none of these

Question

Quizplus · Accepted Answer

Using the chain rule, we have:
$$f'(x) = 8e^x - 3x^2 \cdot \frac{1}{1+x^2} - 2x \cdot \frac{1}{1+x^2}$$
$$ = 8e^x - \frac{3x^2}{1+x^2} - \frac{2x}{1+x^2}$$

Now, to find the second derivative, we take the derivative of the above expression:
$$f''(x) = 8e^x - \frac{(6x+2)x^2}{(1+x^2)^2} + \frac{2(1+x^2)3x^2}{(1+x^2)^2}$$
$$ = 8e^x - \frac{3x^2}{(1+x^2)^2} - \frac{6x}{(1+x^2)^2} + \frac{6x^4}{(1+x^2)^2}$$

Simplifying the above expression, we get:
$$f''(x) = 8e^x - \frac{3x^2}{1+x^2} - \frac{6x}{(1+x^2)^2} + \frac{6x^4}{(1+x^2)^2}$$

Therefore, the answer is D.

Find ft(x)f ^ { t } ( x )ft(x) f(x)=8ex−3x2arctan⁡(x)f ( x ) = 8 e ^ { x } - 3 x ^ { 2 } \arctan ( x )f(x)=8ex−3x2arctan(x)

Find $f ^ { t } ( x )$ $f ( x ) = 8 e ^ { x } - 3 x ^ { 2 } \arctan ( x )$