Find the derivative of the function. $y = \frac { 3 } { 5 } \sec ^ { 2 } x$
A) $y ^ { t } = - \frac { 6 } { 5 } \sec ^ { 2 } x \tan x$
B) $y ^ { t } = \frac { 6 } { 5 } \sec ^ { 2 } x \tan ^ { 2 } x$
C) $y ^ {t } = \frac { 6 } { 5 } \sec x \tan x$
D) $y ^ { t } = \frac { 6 } { 5 } \sec ^ { 2 } x \tan x$
E) $y ^ { t } = \frac { 3 } { 5 } \sec ^ { 2 } x \tan x$

Question

Quizplus · Accepted Answer

To find the derivative of the function, we use the formula:
$$\frac{d}{dx}\sec x=\sec x \tan x$$
So, applying the chain rule, we get:
\begin{align*}
\frac{d}{dx}\left(\frac{3}{5}\sec^2x\right)&=\frac{3}{5}\cdot2\sec x \tan x\
&=\frac{6}{5}\sec x \tan x\
&=\frac{6}{5}\sec^2x\cdot\frac{\sin x}{\cos x}\
&=\frac{6}{5}\cdot\frac{\sin x}{\cos^2x}\cdot\frac{\cos x}{\cos x}\
&=\frac{6}{5}\cdot\frac{\sin x}{\cos^3x}\
&=\frac{6}{5}\cdot\frac{1}{\sec^3x}\cdot\frac{\sin x}{1}\
&=\frac{6}{5}\cdot\frac{1}{\sec^3x}\cdot\tan x
\end{align*}
Therefore, the correct choice is D.

Find the Derivative of the Function y=35sec⁡2xy = \frac { 3 } { 5 } \sec ^ { 2 } xy=53​sec2x

Find the Derivative of the Function $y = \frac { 3 } { 5 } \sec ^ { 2 } x$