$$\text { Find the derivative of the trigonometric function } f ( s ) = s ^ { 3 } \tan s \text {. }$$
A) $f ^ { t } ( s ) = s ^ { 3 } \sec ^ { 2 } s + 3 s ^ { 2 } \tan s$
B) $f ^ { t } ( s ) = s ^ { 3 } \sec ^ { 2 } s + 2 s ^ { 2 } \tan s$
C) $f ^ {t } ( s ) = 3 s ^ { 2 } \tan s - s ^ { 2 } \sec ^ { 2 } s$
D) $f ^ {t } ( s ) = s ^ { 2 } \sec ^ { 2 } s - 3 s ^ { 2 } \tan s$
E) $f ^ { t } ( s ) = s ^ { 3 } \sec s + 3 s ^ { 2 } \tan s$

Question

$$\text { Find the derivative of the trigonometric function } f ( s ) = s ^ { 3 } \tan s \text {. }$$ 
A) $f ^ { t } ( s ) = s ^ { 3 } \sec ^ { 2 } s + 3 s ^ { 2 } \tan s$
B) $f ^ { t } ( s ) = s ^ { 3 } \sec ^ { 2 } s + 2 s ^ { 2 } \tan s$
C) $f ^ {t } ( s ) = 3 s ^ { 2 } \tan s - s ^ { 2 } \sec ^ { 2 } s$
D) $f ^ {t } ( s ) = s ^ { 2 } \sec ^ { 2 } s - 3 s ^ { 2 } \tan s$
E) $f ^ { t } ( s ) = s ^ { 3 } \sec s + 3 s ^ { 2 } \tan s$

Quizplus · Accepted Answer

The derivative of $f(s) = s^3 \tan s$ is found using the product rule, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. Thus, $\frac{d}{ds}(s^3) \cdot \tan s + s^3 \cdot \frac{d}{ds}(\tan s) = 3s^2 \tan s + s^3 \sec^2 s$.

Find the derivative of the trigonometric function f(s)=s3tan⁡s. \text { Find the derivative of the trigonometric function } f ( s ) = s ^ { 3 } \tan s \text {. } Find the derivative of the trigonometric function f(s)=s3tans.

$\text { Find the derivative of the trigonometric function } f ( s ) = s ^ { 3 } \tan s \text {. }$