Question

Differentiate w.r.t $x$,$\sin \sqrt{2x+3}+\tan (7-2x{)}^6$

Answer 1

Let

$y=\sin \sqrt{2x+3}+\tan (7-2x{)}^6$

Differentiate on both sides w.r.t $x$

$\frac{dy}{dx}=\cos \sqrt{2x+3}\frac{d}{dx}\left(\sqrt{2x+3}\right)+{\sec }^2(7-2x{)}^6\frac{d}{dx}((7-2x{)}^6)$

$=\cos \left(\sqrt{2x+3}\right)\frac{2}{2\sqrt{2x+3}}+{\sec }^2(7-2x{)}^6\times 6(7-2x{)}^5\times (-2)$

$\frac{dy}{dx}=\frac{\cos \left(\sqrt{2x+3}\right)}{\sqrt{2x+3}}-12(7-2x{)}^5{\sec }^2(7-2x{)}^6$