Find the derivative.
-$s = t ^ { 3 } \tan t - \sqrt { t }$
A) $\frac { d s } { d t } = t ^ { 3 } \sec t \tan t + 3 t ^ { 2 } \tan t - \frac { 1 } { 2 \sqrt { t } }$
B) $\frac { d s } { d t } = t ^ { 3 } \sec ^ { 2 } t + 3 t ^ { 2 } \tan t - \frac { 1 } { 2 \sqrt { t } }$
C) $\frac { d s } { d t } = - t ^ { 3 } \sec ^ { 2 } t + 3 t ^ { 2 } \tan t + \frac { 1 } { 2 \sqrt { t } }$
D) $\frac { \mathrm { ds } } { \mathrm { dt } } = 3 \mathrm { t } ^ { 2 } \sec ^ { 2 } \mathrm { t } - \frac { 1 } { 2 \sqrt { t } }$

Question

Quizplus · Accepted Answer

Using the product rule and chain rule, we have:

$$\frac{ds}{dt} = (3t^2 	an t + t^3\sec^2t) - \frac{1}{2\sqrt{t}}$$

Since $	an t = \frac{\sin t}{\cos t}$ and $\sec t = \frac{1}{\cos t}$, we can rewrite the first term as:

$$3t^2 	an t + t^3\sec^2t = 3t^2 \cdot \frac{\sin t}{\cos t} + t^3 \cdot \frac{1}{\cos^2t} = t^2\left(3\sin t\cos t + \frac{t}{\cos^2t}ight)$$

Using the identity $\sin 2t = 2\sin t \cos t$, we get:

$$t^2\left(3\sin t\cos t + \frac{t}{\cos^2t}ight) = t^2\left(\frac{3}{2}\sin 2t + \frac{t}{\cos^2t}ight)$$

Finally, using the identity $\cos^2t = 1 - \sin^2t$, we get:

$$t^2\left(\frac{3}{2}\sin 2t + \frac{t}{1 - \sin^2t}ight) - \frac{1}{2\sqrt{t}}$$

Multiplying the second term by $\frac{\sqrt{t}}{\sqrt{t}}$, we get:

$$t^2\left(\frac{3}{2}\sin 2t + \frac{t\sqrt{t}}{t - \sqrt{t}}ight) - \frac{\sqrt{t}}{2(t-\sqrt{t})}$$

This matches the expression in choice B.

Find the Derivative s=t3tan⁡t−ts = t ^ { 3 } \tan t - \sqrt { t }s=t3tant−t​

Find the Derivative $s = t ^ { 3 } \tan t - \sqrt { t }$