Question

Topic - Differentiation

$\boxed{\sf \footnotesize Find \: f'(x) \: when \: f(x) = \pi {e}^{sin(x + y)} - \dfrac{\pi}{4}}$
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Answer 1

• refer attachment for detailed answer

Answer 2

Refer to the attachment for your answer :)

Answer :-

$\quad \qquad \boxed{ \bf \pi \cdot e^{\sin ( x + y )} \cdot \cos(x+y)}$

Used Concepts :-

• $\bf \dfrac{d}{dx} ( Constant ) = 0$

• $\bf \dfrac{d}{dx} ( u \cdot v ) = u \cdot \dfrac{dv}{dx} + v \cdot \dfrac{du}{dx}$

• $\bf \dfrac{d}{dx} \{ \sin ( x ) \} = \cos ( x )$

• Chain Rule of Differentiation

Additional Information :-

• ${ \boxed { \tt { e = \displaystyle \tt \lim_{ \tt n \to \infty } { \bigg ( \tt 1 + \dfrac{1}{n} \bigg )}^{n} }}}$

Maclaurin Series :-

Binomial Expansion $\forall$ x² < 1

• ${ \boxed { \tt { \orange { {( 1 \pm x)}^{n} = 1 \pm \dfrac{nx}{1!} + \dfrac{n(n - 1)x²}{2!} + . . . . . . . }}}}$

• ${ \boxed { \tt { \red { {( 1 \pm x)}^{- n} = 1 \mp \dfrac{nx}{1!} + \dfrac{n(n + 1)x²}{2!} + . . . . . . . . }}}}$

Maclaurin Series of Sin x :-

• ${ \boxed { \tt { \orange { Sin x = x - \dfrac{x³}{3!} + \dfrac{x⁵}{5!} - . . . . . . . . . . }}}}$

Maclaurin Series of Cos x :-

• ${ \boxed { \tt { \green { Cos x = 1 - \dfrac{x²}{2!} + \dfrac{x⁴}{4!} - . . . . . . . . . . }}}}$

Maclaurin Series of tan x :-

• ${ \boxed { \tt { \blue { tan x = x + \dfrac{x³}{3} + \dfrac{2x⁵}{15} + . . . . . . . . . . }}}}$

Maclaurin Series of $\bf e^{x}$ :-

• ${ \boxed { \tt { \green { e^{x} = 1 + x + \dfrac{x²}{2!} + \dfrac{x³}{3!} + . . . . . . . . }}}}$

Maclaurin Series of $\tt {tan}^{ - 1 } x$ If & only $\tt { |x| \leqslant 1 }$ :-

• ${ \boxed { \tt { \orange { {tan}^{-1} x = x - \dfrac{x³}{3} + \dfrac{x⁵}{5} - . . . . . . . . . . }}}}$

Commonly used Derivatives :-

• ${ \boxed { \tt { \red { \dfrac{d}{dx} ( x^{n} ) = n.{(x)^{(n-1)}} }}}}$

• ${ \boxed { \tt { \orange { \dfrac{d}{dx} ( a^{x} ) = a^{x} . log ( a ) }}}}$

• ${ \boxed { \tt { \blue { \dfrac{d}{dx} \bigg ( \dfrac{u}{v} \bigg ) = \dfrac{v . \dfrac{du}{dx} - u . \dfrac{dv}{dx}}{v²} }}}}$

• ${ \boxed { \tt { \green { \dfrac{d}{dx} ( u.v ) = v . \dfrac{du}{dx} + u . \dfrac{dv}{dx} }}}}$