Question

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

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

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