Find derivative of trigonometric function y=(3(1-sin x))/(2cos x)

Question
Trigonometric Functions
asked 2021-02-08
Find derivative of trigonometric function \(\displaystyle{y}=\frac{{{3}{\left({1}-{\sin{{x}}}\right)}}}{{{2}{\cos{{x}}}}}\)

Answers (1)

2021-02-09
Simplify the trigonometric function
\(\displaystyle{y}=\frac{{{3}{\left({1}-{\sin{{x}}}\right)}}}{{{2}{\cos{{x}}}}}\)
\(\displaystyle{y}=\frac{{3}}{{2}}{\left(\frac{{1}}{{\cos{{x}}}}-{\left(\frac{{\sin{{x}}}}{{\cos{{x}}}}\right)}\right)}\)
\(\displaystyle{y}=\frac{{3}}{{2}}{\left({\sec{{x}}}-{\tan{{x}}}\right)}{\left\lbrace\because\frac{{1}}{{{\cos{{x}}}}}={\sec{{x}}}{\quad\text{and}\quad}\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}={\tan{{x}}}\right\rbrace}\)
Derivative of trigonometric function:
\(\displaystyle{y}=\frac{{3}}{{2}}{\left({\sec{{x}}}-{\tan{{x}}}\right)}\)
\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{3}}{{2}}{\left({\sec{{x}}}-{\tan{{x}}}\right)}\right)}\)
\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{3}}{{2}}{\left(\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\sec{{x}}}\right)}-\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\tan{{x}}}\right)}\right)}\)
\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{3}}{{2}}{\left({\left({\sec{{x}}}\cdot{\tan{{x}}}\right)}-{{\sec}^{{2}}{x}}\right)}{\left\lbrace\because\frac{{x}}{{{\left.{d}{x}\right.}}}{\left({\sec{{x}}}\right)}={\left({\sec{{x}}}\cdot{\tan{{x}}}\right)}{\quad\text{and}\quad}{d}{\left({\left.{d}{x}\right.}\right)}{\left({\tan{{x}}}\right)}={{\sec}^{{2}}{x}}\right\rbrace}\)
The derivative of function is given below
\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{3}}{{2}}{\left({\left({\sec{{x}}}\cdot{\tan{{x}}}\right)}-{{\sec}^{{2}}{x}}\right)}\)
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