Question

find the derivative of √tan 3z (whole root over of tan3z)​

Answer 1

derivation wrt z.........

Answer 2

Answer:

The derivative is $\frac{3 Sec^{2}3z }{2 tan3z}$

Explanation:

Given: $\sqrt{tan3z}$

Find: The derivative

Solution: according to the question.

$y=\sqrt{tan3z}$

$\frac{dy}{d_2}=\frac {d}{d_2} \sqrt{tan3z}$

                               $\therefore \frac{d\sqrt{x} }{dx}=\frac{1}{2\sqrt{x}} }$

$=\frac{1}{2\sqrt{tan3z} } \times \frac {d}{dz}(tan32)$

                               $\because dtanx=sec^2x$

$= \frac{1}{2\sqrt{tan3z}}\times sec^232\times \frac {d}{d_2}(3z)$

$\frac {sec^232}{2\sqrt{tan3z} }\times \frac{3dz}{dz}$

$=\frac{3 sec^23z}{2tan3z}$

Hence, the derivative is $\frac{3}{2} \frac{sec^2 3z}{tan3z}$

#SPJ2