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Question: derivative of sin inverse x by first principle...

derivative of sin inverse x by first principle

Answer

11−x2\frac{1}{\sqrt{1-x^2}}

Explanation

Solution

To find the derivative of f(x)=sin⁡−1xf(x) = \sin^{-1}x using the first principle, we use the definition: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} Substitute f(x)=sin⁡−1xf(x) = \sin^{-1}x: f′(x)=lim⁡h→0sin⁡−1(x+h)−sin⁡−1xhf'(x) = \lim_{h \to 0} \frac{\sin^{-1}(x+h) - \sin^{-1}x}{h} Let y=sin⁡−1xy = \sin^{-1}x and y+k=sin⁡−1(x+h)y+k = \sin^{-1}(x+h).

From these substitutions, we have: x=sin⁡yx = \sin y x+h=sin⁡(y+k)x+h = \sin(y+k)

Subtracting the first equation from the second gives: h=sin⁡(y+k)−sin⁡yh = \sin(y+k) - \sin y

Also, as h→0h \to 0, x+h→xx+h \to x. Since sin⁡−1x\sin^{-1}x is a continuous function, sin⁡−1(x+h)→sin⁡−1x\sin^{-1}(x+h) \to \sin^{-1}x, which means y+k→yy+k \to y. Therefore, as h→0h \to 0, k→0k \to 0.

Now, substitute these expressions back into the limit: f′(x)=lim⁡k→0(y+k)−ysin⁡(y+k)−sin⁡yf'(x) = \lim_{k \to 0} \frac{(y+k) - y}{\sin(y+k) - \sin y} f′(x)=lim⁡k→0ksin⁡(y+k)−sin⁡yf'(x) = \lim_{k \to 0} \frac{k}{\sin(y+k) - \sin y}

Use the trigonometric identity sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right). Here, A=y+kA = y+k and B=yB = y. So, sin⁡(y+k)−sin⁡y=2cos⁡(y+k2)sin⁡(k2)\sin(y+k) - \sin y = 2 \cos\left(y+\frac{k}{2}\right) \sin\left(\frac{k}{2}\right) =2cos⁡(y+k2)sin⁡(k2)= 2 \cos\left(y+\frac{k}{2}\right) \sin\left(\frac{k}{2}\right)

Substitute this back into the limit expression: f′(x)=lim⁡k→0k2cos⁡(y+k2)sin⁡(k2)f'(x) = \lim_{k \to 0} \frac{k}{2 \cos\left(y+\frac{k}{2}\right) \sin\left(\frac{k}{2}\right)}

Rearrange the terms to use the standard limit lim⁡θ→0sin⁡θθ=1\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1: f′(x)=lim⁡k→012cos⁡(y+k2)⋅ksin⁡(k2)f'(x) = \lim_{k \to 0} \frac{1}{2 \cos\left(y+\frac{k}{2}\right)} \cdot \frac{k}{\sin\left(\frac{k}{2}\right)} As k→0k \to 0:

The first part of the product becomes: lim⁡k→012cos⁡(y+k2)=12cos⁡y\lim_{k \to 0} \frac{1}{2 \cos\left(y+\frac{k}{2}\right)} = \frac{1}{2 \cos y} The second part of the product becomes: lim⁡k→0ksin⁡(k2)=lim⁡k→02⋅(k/2)sin⁡(k/2)\lim_{k \to 0} \frac{k}{\sin\left(\frac{k}{2}\right)} = \lim_{k \to 0} \frac{2 \cdot (k/2)}{\sin(k/2)}

Let θ=k/2\theta = k/2. As k→0k \to 0, θ→0\theta \to 0. So, lim⁡θ→02θsin⁡θ=2lim⁡θ→0θsin⁡θ=2⋅1=2\lim_{\theta \to 0} \frac{2\theta}{\sin\theta} = 2 \lim_{\theta \to 0} \frac{\theta}{\sin\theta} = 2 \cdot 1 = 2.

Combine these results: f′(x)=12cos⁡y⋅2=1cos⁡yf'(x) = \frac{1}{2 \cos y} \cdot 2 = \frac{1}{\cos y}

Now, we need to express cos⁡y\cos y in terms of xx. We know that y=sin⁡−1xy = \sin^{-1}x, which implies sin⁡y=x\sin y = x. Using the identity sin⁡2y+cos⁡2y=1\sin^2 y + \cos^2 y = 1, we get: cos⁡2y=1−sin⁡2y\cos^2 y = 1 - \sin^2 y cos⁡2y=1−x2\cos^2 y = 1 - x^2 cos⁡y=±1−x2\cos y = \pm \sqrt{1 - x^2}

The principal value branch of sin⁡−1x\sin^{-1}x is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]. In this interval, cos⁡y\cos y is non-negative. Therefore, cos⁡y=1−x2\cos y = \sqrt{1 - x^2}.

Substitute this back into the expression for f′(x)f'(x): f′(x)=11−x2f'(x) = \frac{1}{\sqrt{1 - x^2}}

The derivative exists for x∈(−1,1)x \in (-1, 1).