7.3.3 Product and Quotient of Functions

Find the derivative of $f:ℝ\to ℝ$ with $f(x)=x^2·e{}^x$. The product rule can be applied choosing, for example, $u(x)=x^2$ and $v(x)=e{}^x$. The corresponding derivatives are $u\text{'}(x)=2x$ and $v\text{'}(x)=e{}^x$. Combining these terms according to the product rule results in the derivative of the function $f$:

${\displaystyle f\text{'}:ℝ\to ℝ\hspace{0.5em},\hspace{0.5em}\hspace{0.5em}x\to f\text{'}(x)=2xe{}^x+x^{2}e{}^x=(x^2+2x)e{}^x\hspace{0.5em}.}$

Next, we investigate the tangent function $g$ with $g(x)=\tan (x)=\frac{\sin (x)}{\cos (x)}$ ($\cos (x)\ne 0$).In order to use the quotient rule we set $u(x)=\sin (x)$ and $v(x)=\cos (x)$. The corresponding derivatives are $u\text{'}(x)=\cos (x)$ and $v\text{'}(x)=-\sin (x)$. Combining these terms and applying the quotient rule results in the derivative of the function $g$:

${\displaystyle g\text{'}(x)=\frac{\cos (x)·\cos (x)-\sin (x)·(-\sin (x))}{{\cos }^2(x)}\hspace{0.5em}.}$
This result can be transformed into any of the following expressions:

${\displaystyle g\text{'}(x)=1+{\left(\frac{\sin (x)}{\cos (x)}\right)}^2=1+{\tan }^2(x)=\frac{1}{{\cos }^2(x)}\hspace{0.5em}.}$
For the last transformation, the relation ${\sin }^2(x)+{\cos }^2(x)=1$ was used, which was given in Module 5 (see Section 5.6.2).