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Derive the differentiation formula \(\dfrac{d}{d x}[\arcsin x]=\dfrac{1}{\sqrt{1-x^{2}}}\).
in Mathematics by Diamond (45,410 points) | 242 views

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To derive the differentiation formula for \(\frac{d}{d x}[\arcsin x]=\frac{1}{\sqrt{1-x^2}}\), we will start by defining \(y=\arcsin x\). This means that \(x=\sin y\)

Now, we'll differentiate both sides of the equation \(x=\sin y\) with respect to \(x\), applying implicit differentiation. Remember that we'll need to use the chain rule when differentiating the term involving \(y\) :
\frac{d}{d x}(x)=\frac{d}{d x}(\sin y)
Differentiating both sides, we get:
1=(\cos y) \frac{d y}{d x}
Now, we need to express \(\cos y\) in terms of \(x\). To do this, we use the Pythagorean identity:
\sin ^2 y+\cos ^2 y=1
Since we know that \(x=\sin y\), we can substitute \(x\) for \(\sin y\) :
x^2+\cos ^2 y=1
Now, solve for \(\cos y\) :

\cos y=\sqrt{1-x^2}
Substitute this expression for \(\cos y\) back into our original equation:
1=\left(\sqrt{1-x^2}\right) \frac{d y}{d x}
Now, solve for \(\frac{d y}{d x}\) :
\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}
Since \(y=\arcsin x\), we can express the result as:
\frac{d}{d x}[\arcsin x]=\frac{1}{\sqrt{1-x^2}}
And we've derived the differentiation formula for the arcsine function.
by Diamond (89,043 points)

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