Question

Find the derivative of $f\left(x\right)=\frac{x\cos x}{\sqrt{1+x^2}}$.

Answer 1

$\begin{array}{c}wehave\\ f\left(x\right)=\frac{x\cos x}{\sqrt{1+x^2}}\\ Putx=\tan t\\ dx={\sec }^{2}dt\\ f\left(\tan t\right)=\frac{\tan t\cos \left(tant\right)}{\sec t}\\ f^{\prime }\left(x\right)dx=f^{\prime }\left(\tan t\right){\sec }^{2}tdt=d\left(\sin t\cos \left(\tan t\right)\right)\\ =\left(\cos t\cos \left(\tan t\right)-\sin t\sin \left(\tan t\right){\sec }^{2}t\right)dt\\ =\left[{\cos }^{3}t\cos \left(\tan t\right)-\sin t\sin \left(\tan t\right)\right]dx\\ f^{\prime }\left(x\right)dx=\left[{\cos }^{3}t\cos \left(\tan t\right)-\sin t\sin \left(\tan t\right)\right]dx\\ =\left[{\left(\frac{1}{\sqrt{1+x^2}}\right)}^3\cos x-\frac{x}{\sqrt{1+x^2}}\sin x\right]dx\\ \frac{d}{dx}\left[f\left(x\right)\right]=\left[\frac{\cos x}{{\left(1+x^2\right)}^{\frac{3}{2}}}-\frac{x\sin x}{{\left(1+x^2\right)}^{\frac{1}{2}}}\right]\end{array}$