Derivative of sine inverse x. In this article, we will find the derivative of sin inverse x using the substitution method and the limit definition. In the end, we will evaluate the derivative of sine inverse 1/x as an application.
Table of Contents
What is the Derivative of Sin Inverse x?
(Substitution Method)
Step 1: Let y=sin-1 x
Step 2: Applying the sine function on both sides, we have
sin y =sin sin-1 x
⇒ sin y =x $\cdots (i)$
Step 3: Differentiating with respect to x, we get
$\cos y \dfrac{dy}{dx}=\dfrac{d}{dx}(x)$
$\Rightarrow \cos y \dfrac{dy}{dx}=1$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos y}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-\sin^2y}}$ $[\because \sin^2y+\cos^2y=1]$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}$ $[\because \sin y =x \text{ from}$ $(i)]$
So the derivative of sine inverse x is $1/\sqrt{1-x^2}$. That is,
\[\dfrac{d}{dx}(\sin^{-1} x)=\dfrac{1}{\sqrt{1-x^2}}. \quad ♣\]
Now, we will find the derivative of sin-1 x by the first principle.
Derivative of sin inverse x from first principle
Let $f(x)=\sin^{-1}x.$ The derivative of $f(x)$ using first principle of derivatives is given below:
$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
Thus the derivative of sine inverse x by first principle of derivatives is
$\dfrac{d}{dx}(\sin^{-1}x)$ $= \lim\limits_{h \to 0} \dfrac{\sin^{-1}(x+h)-\sin^{-1}x}{h}$ $\cdots (ii)$
Suppose that $\sin^{-1}(x+h)= \alpha$ and $\sin^{-1}(x)=\beta$
⇒ $x+h=\sin \alpha$ and $x=\sin \beta$
Also, $h =(x+h)-h$ $=\sin \beta-\sin\alpha.$ As $h \to 0$ we have $\sin\alpha \to \sin \beta$. This implies that $\alpha \to \beta$ for $-1 \leq \alpha\leq 1.$ Thus from (ii) we get that
$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{\alpha \to \beta}\dfrac{\alpha – \beta}{\sin \alpha -\sin \beta}$
$=\lim\limits_{\alpha\to \beta} \dfrac{\alpha-\beta}{2\cos(\frac{\alpha+\beta}{2})\sin(\frac{\alpha-\beta}{2})}$
$=\lim\limits_{\alpha\to \beta} \dfrac{\frac{\alpha-\beta}{2}}{\cos(\frac{\alpha+\beta}{2})\sin(\frac{\alpha-\beta}{2})}$
$=\lim\limits_{\alpha\to \beta} \dfrac{1}{\cos(\frac{\alpha+\beta}{2})} \times$ $\lim\limits_{\alpha\to \beta} \dfrac{\frac{\alpha-\beta}{2}}{\sin(\frac{\alpha-\beta}{2})}$
[Let $\frac{\alpha-\beta}{2}=t.$ Then $t \to 0$ as $\alpha \to \beta$]
$=\lim\limits_{\alpha\to \beta} \dfrac{1}{\cos(\frac{\alpha+\beta}{2})} \times$ $\lim\limits_{t \to 0} \dfrac{t}{\sin t}$
$=\dfrac{1}{\cos \beta} \times 1$
$=\dfrac{1}{\sqrt{1-x^2}}$ as we know that $\cos \beta$ $=\cos \sin^{1} x$ $=\cos \cos^{-1} \sqrt{1-x^2}$ $=\sqrt{1-x^2}$
This shows that the derivative of sine inverse x is $\frac{1}{\sqrt{1-x^2}}.$ ♣
Also Read:
Derivative of root x: The derivative of √x is 1/2√x
Derivative of cube root of x: The derivative of the cube root of x is 1/(3x^{2/3})
Integration of root x: The integration of √x is 2/3x^{3/2}
Integration of mod x: The integration of |x| is -x|x|/2 +c
In the above, we have calculated the derivative of sine inverse x which is
$\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}$ $\cdots (*)$
As an application of the above formula, we will now find the derivative of sine inverse 1/x.
Derivative of Sin Inverse 1/x by chain rule
Now, using the chain rule we will find the derivative of sine inverse 1/x. The chain rule of derivatives is given below: $$\frac{du}{dx}=\frac{du}{dz} \cdot \frac{dz}{dx}.$$
To find the derivative of $\sin^{-1}(\frac{1}{x})$ by the chain rule, let us put
$z=\dfrac{1}{x}$
Now, $\dfrac{d}{dx}(\sin^{-1}(\frac{1}{x}))=\dfrac{d}{dx}(\sin^{-1}z)$
$=\dfrac{d}{dz}(\sin^{-1}z) \cdot \dfrac{dz}{dx}$, by the chain rule.
$=\dfrac{d}{dz}(\sin^{-1}z) \cdot \dfrac{d}{dx}(\dfrac{1}{x})$
[Now, $\frac{1}{x}=x^{-1}.$ So by the power rule of derivatives, we have $\frac{d}{dx}(\frac{1}{x})$ $=\frac{d}{dx}(x^{-1})$ $=-x^{-2}$ $=-\dfrac{1}{x^2}.$]
$=\dfrac{1}{\sqrt{1-z^2}} \cdot (-\dfrac{1}{x^2})$
$=-\dfrac{1}{\sqrt{1-(\frac{1}{x})^2}} \cdot \dfrac{1}{x^2}$
$=-\dfrac{1}{\sqrt{\frac{x^2-1}{x^2}}} \cdot \dfrac{1}{x^2}$
$=-\dfrac{\sqrt{x^2}}{\sqrt{x^2-1}} \cdot \dfrac{1}{x^2}$
$=-\dfrac{1}{x\sqrt{x^2-1}}$
