Question

derivative of sec^2x + tan^2x​

Answer 1

EXPLANATION.

Derivative of :

⇒ sec²x + tan²x.

As we know that,

⇒ d(sec x)/dx = sec(x)tan(x).

⇒ d(tan x)/dx = sec²x.

Using this formula in equation, we get.

By applying chain rule method, we get.

⇒ y = sec²x + tan²x.

⇒ dy/dx = 2 sec(x).d(sec x)/dx + 2 tan(x).d(tan x)/dx.

⇒ dy/dx = 2 sec(x).sec(x).tan(x) + 2 tan(x).sec²(x).

⇒ dy/dx = 2sec²x.tan(x) + 2tan(x).sec²(x).

⇒ dy/dx = 4sec²x.tan(x).

                                                                                                                         

MORE INFORMATION.

Nth derivatives of some standard functions.

$\sf (1) = \dfrac{d^{n} }{dx^{n} } sin(ax + b) = a^{n} sin \bigg(\dfrac{n \pi}{2} + ax + b \bigg).$

$\sf (2) = \dfrac{d^{n} }{dx^{n} } cos(ax + b) = a^{n} cos \bigg(\dfrac{n \pi}{2} + ax + b \bigg).$

$\sf (3) = \dfrac{d^{n} }{dx^{n} }(ax + b)^{m} = \dfrac{m !}{(m - n)!} a^{n} (ax + b)^{m - n} , where \ m > n$

$\sf (4) = \dfrac{d^{n} }{dx^{n} } (log(ax + b)) = \dfrac{(-1)^{n - 1} (n - 1)! a^{n} }{(ax + b)^{n} }$

$\sf (5) = \dfrac{d^{n} }{dx^{n} }(e^{ax} ) = a^{n} e^{ax} .$

$\sf (6) = \dfrac{d^{n} (a^{x} ) }{dx^{n} } = a^{x} (log \ a)^{n}$

$\sf (7) = \dfrac{d^{n} }{dx^{n} } (e^{ax} sin(bx + c)) = r^{n} e^{ax} sin(bx + c + n \phi) \\ \\ where \ r = \sqrt{a^{2} + b^{2} } \ ; \ \phi = tan^{-1} \bigg(\dfrac{b}{a}\bigg).$

$\sf (8) = \dfrac{d^{n} }{dx^{n} } (e^{ax} cos(bx + c)) = r^{n} e^{ax} cos(bx + c + n \phi) \\ \\ where \ r = \sqrt{a^{2} + b^{2} } \ ; \phi = tan^{-1} \bigg(\dfrac{b}{a} \bigg).$