Question

Find the derivatives of the given functions. w(x)=\sec x\tan (x^{2}-1)

Derivatives
ANSWERED
asked 2021-06-19
Find the derivatives of the given functions. \(\displaystyle{w}{\left({x}\right)}={\sec{{x}}}{\tan{{\left({x}^{{{2}}}-{1}\right)}}}\)

Answers (1)

2021-06-20
\(\displaystyle{w}{\left({x}\right)}={\sec{{x}}}{\tan{{\left({x}^{{{2}}}-{1}\right)}}}\)
\(\displaystyle{w}'{\left({x}\right)}={\sec{{x}}}\dot{{\lbrace}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\tan{{\left({x}^{{{2}}}-{1}\right)}}}+{\tan{{\left({x}^{{{2}}}-{1}\right)}}}\dot{{\lbrace}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\sec{{x}}}\right)}\right.}\)
\(\displaystyle={\sec{{x}}}\dot{{\lbrace}}{\left({x}^{{{2}}}-{1}\right)}\dot{{\lbrace}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}^{{{2}}}-{1}\right)}+{\tan{{\left({x}^{{{2}}}-{1}\right)}}}\dot{{\lbrace}}{\sec{{x}}}\dot{{\tan{{x}}}}\)
\(\displaystyle=={\sec{{x}}}\dot{{\lbrace}}{{\sec}^{{{2}}}{\left({x}^{{{2}}}-{1}\right)}}\dot{{\lbrace}}{2}{x}+{\sec{{x}}}\dot{{\lbrace}}{\tan{{x}}}\dot{{\lbrace}}{\tan{{\left({x}^{{{2}}}-{1}\right)}}}\)
\(\displaystyle={2}{x}\dot{{\lbrace}}{\sec{{x}}}\dot{{\lbrace}}{{\sec}^{{{2}}}{\left({x}^{{{2}}}-{1}\right)}}+{\sec{{x}}}\dot{{\lbrace}}{\tan{{x}}}\dot{{\lbrace}}{\tan{{\left({x}^{{{2}}}-{1}\right)}}}\)
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