反三角函数的微分

在本文中，我们将探索隐式微分的应用，以找到反三角函数的导数。但是在前进之前，让我们回顾一下隐式微分和逆三角学的概念。

反三角学

三角函数的反函数是三角比的反函数，即sin，cos，tan，cot，sec，cosec。这些功能广泛用于物理，数学，工程和其他研究领域。就像加法和减法是彼此的反函数一样，三角函数的反函数也是如此。

sin θ = x

⇒ θ ​= sin⁻¹x​
 

反三角函数的表示

它们通过在前缀中添加弧线或在功率上加上-1来表示。

Inverse sine can be written in two ways:

• sin^-1 x
• arcsin x

Same goes for cos and tan.

Note: Don’t confuse sin^-1 x with (sin x)^-1. They are different. Writing sin^-1 x is a way to write inverse sine whereas (sin x)^-1 means 1/sin x.

隐分化

隐式微分是一种利用链规则来区分隐式定义的函数的方法。显式地找到函数然后进行区分通常并不容易。相反，我们可以完全微分f(x，y)，然后求解方程的其余部分以找到f’(x)的值。即使可以明确求解原始方程式，总的来说，由全微分得出的公式也更简单易用。让我们区分一些反三角函数。

示例1：区分正弦^-1 (x)?

解决方案：

Let, 

y = sin⁻¹ (x)

Taking sine on both sides of equation gives,

$sin (y) = sin (sin^{-1}(x))$

By the property of inverse trigonometry we know, $sin (sin^{-1}(x))=x$

$sin (y) = x$

Now differentiating both sides wrt to x,

$\frac{d}{dx}sin (y)= \frac{d}{dx}x$

$cos(y). \frac{dy}{dx} = 1$

 $\frac{dy}{dx} = \frac{1} { cos(y)}$

We can simplify it more by using the below observation:

$\begin{aligned} sin^2(y)+cos^2(y)&=1 \\x^2+cos^2(y)&=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( As\ sin (y) = x )\\cos^2(y)&=1-x^2\\ cos(y)&=\sqrt{1-x^2}\\ \end{aligned}$

Substituting the value, we get

$\begin{aligned} \frac{dy}{dx}&=\frac{1}{cos(y)} \\ \frac{dy}{dx}&=\frac{1}{\sqrt{1-x^2}} \end{aligned}$

示例2：区分cos ^-1 (x)?

解决方案：

Let, 

y = cos⁻¹ (x)

Taking cosine on both sides of equation gives,

$cos (y) = cos (cos^{-1}(x))$

By the property of inverse trigonometry we know, $cos (cos^{-1}(x))=x$

$cos (y) = x$

Now differentiating both sides wrt to x,

$\frac{d}{dx}cos (y)= \frac{d}{dx}x$

$-sin(y). \frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{-1} { sin(y)}$

We can simplify it more by using the below observation:

$\begin{aligned} sin^2(y)+cos^2(y)&=1 \\sin^2(y)+x^2&=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( As\ cos (y) = x )\\sin^2(y)&=1-x^2\\ sin(y)&=\sqrt{1-x^2}\\ \end{aligned}$

Substituting the value, we get

$\begin{aligned} \frac{dy}{dx}&=\frac{-1}{sin(y)} \\ \frac{dy}{dx}&=\frac{-1}{\sqrt{1-x^2}} \end{aligned}$

示例3：区分tan ^-1 (x)?

解决方案：

Let, 

y = tan⁻¹ (x)

Taking tan on both sides of equation gives,

$tan (y) = tan (tan^{-1}(x))$

By the property of inverse trigonometry we know,$tan (tan^{-1}(x))=x$

$tan (y) = x$

Now differentiating both sides wrt to x,

$\frac{d}{dx}tan (y)= \frac{d}{dx}x$

$sec^2(x). \frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1} {sec^2(x)}$

We can simplify it more by using the below observation:

$\begin{aligned} sec^2(y)-tan^2(y)&=1 \\sec^2(y)-x^2&=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( As\ tan (y) = x )\\sec^2(y)&=1+x^2\end{aligned}$

Substituting the value, we get

$\begin{aligned} \frac{dy}{dx}&=\frac{1}{sec^2(y)} \\ \frac{dy}{dx}&=\frac{1}{{1+x^2}} \end{aligned}$

反三角函数微分的一些高级示例

示例1：y = cos ^-1 (-2x ² )。在y = 1/2处找到dy / dx?

解决方案：

方法1(使用隐式微分)

Given, 

          y = cos⁻¹ (−2x²)

⇒ cos y = −2x²

Differentiating both sides wrt x

$\begin{aligned} \frac{d}{dx} cos(y) &=\frac{d}{dx} -2x^2 \\ -sin(y) \frac{dy}{dx} &=-4x \\ \frac{dy}{dx}&=\frac{4x}{sin (y)} \end{aligned}$

Simplifying

$\begin{aligned} sin^2(y)+cos^2(y)&=1 \\sin^2(y)+(-2x^2)^2 &=1 \\sin^2(y)+4x^4 &=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( As\ cos (y) = -2x^2 )\\sin^2(y)&=1-4x^4\\ sin(y)&=\sqrt{1-4x^4}\\ \end{aligned}$

Putting the obtained value we get,

$\begin{aligned} \frac{dy}{dx} &=\frac{4x}{\sqrt{1-4x^4}} \\ &=\frac{4(\frac{1}{2})}{\sqrt{1-4(\frac{1}{2})^4}} \\ &=\frac{2}{\sqrt{1-\frac{1}{4}}} \\ &=\frac{4}{\sqrt{3}} \end{aligned}$

方法2(使用链规则，因为我们知道arccos x的微分)

Given, 

          y = cos⁻¹ (−2x²)

Differentiating both sides wrt x

$\begin{aligned} \frac{dy}{dx} &=\frac{d}{dx} cos^{-1}(-2x^2) \\ &=\frac{-1}{\sqrt{1-(-2x^2)^2}}\ .\ (-4x) \\ &=\frac{4x}{\sqrt{1-4x^4}} \\ &=\frac{4(\frac{1}{2})}{\sqrt{1-4(\frac{1}{2})^4}} \\ &=\frac{2}{\sqrt{1-\frac{1}{4}}} \\ &=\frac{4}{\sqrt{3}} \end{aligned}$
 

示例2：区分$\begin{aligned}sin^{-1}(\frac{1-x}{1+x}) \end{aligned}$ ?

Let, 

$\begin{aligned} y = sin^{-1}(\frac{1-x}{1+x}) \end{aligned}$

Differentiating both sides wrt x

$\begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}sin^{-1}(\frac{1-x}{1+x}) \\ &= \frac{1}{\sqrt{1-(\frac{1-x}{1+x})^2}} \ . \frac{d}{dx}(\frac{1-x}{1+x}) \\ &= \frac{1+x}{\sqrt{(1+x)^2-({1-x})^2}} \ . \frac{-(1+x)-(1-x)}{(1+x)^2} \\ &= \frac{1}{\sqrt{(1+x)^2-({1-x})^2}} \ . \frac{-2}{(1+x)} \\ &= \frac{1}{\sqrt{4x}} \ . \frac{-2}{(1+x)} \\ &= \frac{-1}{\sqrt{x}(1+x)} \\ \end{aligned}$

逆三角函数的导数表

Function	Derivative
$sin^{-1} x$	$\frac{1}{\sqrt{1-x^2}}$
$cos^{-1} x$	$\frac{-1}{\sqrt{1-x^2}}$
$tan^{-1} x$	$\frac{1}{{1+x^2}}$
$cot^{-1} x$	$\frac{-1}{{1+x^2}}$
$sec^{-1} x$	$\frac{1}{|x|\sqrt{x^2-1}}$
$cosec^{-1} x$	$\frac{-1}{|x|\sqrt{x^2-1}}$