Question

Prove that d/dx(x/2 root of a2-x2 + a2/2 sin-1x/a) = root of a2-x2

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$&\frac{d}{d x}\left\{\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right\}=\sqrt{a^{2}-x^{2}}$ Hence it is proved.

Step-by-step explanation:

$&\frac{d}{d x}\left\{\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right\}=\sqrt{a^{2}-x^{2}}$

$&\text { LHS }=\frac{d}{d x}\left\{\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right\}$

$&=\frac{d}{d x}\left(\frac{x}{2} \sqrt{a^{2}-x^{2}}\right)+\frac{d}{d x}\left(\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right)$

$&=\frac{1}{2}\left[x \frac{d}{d x} \sqrt{a^{2}-x^{2}}+\sqrt{a^{2}-x^{2}} \frac{d}{d x}(x)\right]+\frac{a^{2}}{2} \times \frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^{2}} \times \frac{d}{d x}}\left(\frac{x}{a}\right)$

$=\frac{1}{2}\leftx \times \frac{d}{2 \sqrt{a^{2}-x^{2}}} \frac{d x}{\left.\left.d a^{2}-x^{2}\right)+\sqrt{a^{2}-x^{2}}\right]+\left[\frac{a^{2}}{2}\right] \times \frac{1}{\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}} \times\left(\frac{1}{a}\right)}\right.$

$&=\frac{1}{2}\left[\frac{-2 x^{2}+2\left(a^{2}-x^{2}\right)}{2 \sqrt{a^{2}-x^{2}}}\right]+\frac{a^{2}}{2 \sqrt{a^{2}-x^{2}}}$

$&=\frac{1}{2}\left[\frac{2\left(a^{2}-2 x^{2}\right)}{2 \sqrt{a^{2}-x^{2}}}\right]+\frac{a^{2}}{2 \sqrt{a^{2}-x^{2}}}$

$&=\frac{a^{2}-2 x^{2}}{2 \sqrt{a^{2}-x^{2}}}+\frac{a^{2}}{2 \sqrt{a^{2}-x^{2}}}$

$&=\frac{a^{2}-2 x^{2}+a^{2}}{2 \sqrt{a^{2}-x^{2}}}$

$&=\frac{2 a^{2}-2 x^{2}}{2 \sqrt{a^{2}-x^{2}}}$

$&=\frac{2 (a^{2}- x^{2})}{2 \sqrt{a^{2}-x^{2}}}\\&=\frac{ (a^{2}- x^{2})}{ \sqrt{a^{2}-x^{2}}}\\&={ \sqrt{a^{2}-x^{2}}}=RHS$