Derivative of arcsin x f′ f(x)=arcsinx ∀x∈]–1,1[,f′(x)=11−x2 Proof Remember that functionarcsinis the inverse function ofsin: (f−1∘f)=(sin∘arcsin)(x)=sin(arcsin(x))=x Using result ofderivative of inverse functions, we have: ...
Derivative $f’$ of function $f(x)=\arcsin{x}$ is: \(\forall x \in ]–1, 1[ ,\quad f'(x) = \dfrac{1}{\sqrt{1-x^2}}\) Proof Remember that function $\arcsin$ is the inverse function of $\sin$ : \[\left(f^{-1} \circ f\right)=\left(\sin \circ \arcsin\right)(...
The inverse sine function is one of the inversetrigonometric functionswhich determines the inverse of the sine function and is denoted as sin-1or Arcsine. For example: If thevalueof sine 90 degree is 1, then the value of inverse sin 1 or sin-1(1) will be equal to 90°. Each trigonom...
Derivativef′of the functionf(x)=sinxis:∀x∈]−∞,+∞[,f′(x)=cosx Proof/Demonstration sin(x+h)−sinxh=sin(x)cos(h)+cos(x)sin(h)−sinxhsin(x+h)−sinxh=sinhh×cosx+sinx×coshh−sinx×1hsin(x+h)...
to wikipedia: Can someone explain to me with a mathematical proof the following: $$ \frac {\partial f(x)} {\partial v} = \hat v \cdot \nabla f(x) $$ I don't get this identity except the special example where the partial derivative of f(x) wrt x is a special kind of a......
Proof/Demonstration By applying the Chain Rulederivative of composite functionavec $u=f,v=f^{-1}$, we have: \(\left(f \circ f^{-1}\right)^{\prime}(x)=f^{\prime}(f^{-1}(x)) \cdot (f^{-1})^{\prime}(x)\) However, by definition of a reciprocal function: ...
Derivative $f’$ of the function $f(x)=\exp x= e^{x}$ is: \(\forall x \in ]-\infty, +\infty[ , f'(x) = \exp x = e^{x}\) Proof/Demonstration \[\begin{aligned} f^\prime(x)=(e^x)^\prime &=\lim _{h \rightarrow 0} \frac{e^{x+h}-e^{x}}{h} \\ &=\lim...
Derivative $f’$ of function $f(x)=\arccos{x}$ is: \[\forall x \in ]–1, 1[ ,\quad f'(x) = -\frac{1}{\sqrt{1-x^2}}\] Proof Remember that function $\arcsin$ is the inverse function of $\cos$ : \[\left(f^{-1} \circ f\right)=\left(\cos \circ \arccos\right)...
Derivative $f’$ of function $f(x)=\arctan{x}$ is: \(\forall x \in \mathbb{R} ,\quad f'(x) = \dfrac{1}{1+x^2}\) Proof Remember that function $\arctan$ is the inverse function of $\tan$ : \[\left(f^{-1} \circ f\right)=\left(\tan \circ \arctan\right)(x)=\...
Derivative $f’$ of the function $f(x)=\tan x$ is: \(\forall x \neq \frac{\pi}{2}+k\pi, k \in \mathbb{Z}, f'(x) = 1+\tan ^{2} x\) Proof First we have: \((\tan x)' =\lim _{h \rightarrow 0} \dfrac{\tan (x+h) - \tan x }{h}\) ...