nd the derivative of the function. y = 3tan⁻¹(x - √1 + x²)

Explanation:

Step1: Recall derivative formula

The derivative of $y = a\tan^{- 1}(u)$ is $y'=\frac{a}{1 + u^{2}}\cdot u'$, where $a = 3$ and $u=x-\sqrt{1 + x^{2}}$.

Step2: Find derivative of $u$

The derivative of $x$ is $1$. For $v=\sqrt{1 + x^{2}}=(1 + x^{2})^{\frac{1}{2}}$, by the chain - rule, $v'=\frac{1}{2}(1 + x^{2})^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{1 + x^{2}}}$. So $u'=1-\frac{x}{\sqrt{1 + x^{2}}}=\frac{\sqrt{1 + x^{2}}-x}{\sqrt{1 + x^{2}}}$.

Step3: Calculate $y'$

$y'=\frac{3}{1+(x - \sqrt{1 + x^{2}})^{2}}\cdot\frac{\sqrt{1 + x^{2}}-x}{\sqrt{1 + x^{2}}}$. First, expand $(x-\sqrt{1 + x^{2}})^{2}=x^{2}-2x\sqrt{1 + x^{2}}+1 + x^{2}=2x^{2}+1-2x\sqrt{1 + x^{2}}$. Then $1+(x - \sqrt{1 + x^{2}})^{2}=1+2x^{2}+1-2x\sqrt{1 + x^{2}}=2x^{2}+2-2x\sqrt{1 + x^{2}}=2(x^{2}+1 - x\sqrt{1 + x^{2}})$. $y'=\frac{3(\sqrt{1 + x^{2}}-x)}{2\sqrt{1 + x^{2}}(x^{2}+1 - x\sqrt{1 + x^{2}})}$. Rationalize the denominator: Multiply the numerator and denominator by $\sqrt{1 + x^{2}}+x$. The denominator becomes $2\sqrt{1 + x^{2}}(x^{2}+1 - x\sqrt{1 + x^{2}})(\sqrt{1 + x^{2}}+x)$. The numerator is $3(\sqrt{1 + x^{2}}-x)(\sqrt{1 + x^{2}}+x)=3((1 + x^{2})-x^{2}) = 3$. After simplification, $y'=\frac{3}{2\sqrt{1 + x^{2}}}$.

Answer:

$\frac{3}{2\sqrt{1 + x^{2}}}$