Question

how to differentiate this wrt t
(√2a+2ut).[√(a²-(a/√2+ut)²)]
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.plzz solve this

Answer 1

Explanation:

$\sf{\frac{d}{dt}[(\sqrt{2}a+ut).(\sqrt{a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}}\:)]}$

Using chain rule,,

$\boxed{\sf{\frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)}}$

We get,,

$\sf{\frac{d}{dt}[(\sqrt{2}a+ut).(\sqrt{a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}}\:) ]}$

$\sf{=(\sqrt{2}a+ut)\frac{d}{dt}(\sqrt{a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}}\:)+(\sqrt{a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}}\:)\frac{d}{dt}(\sqrt{2}a+ut)}$

$\sf{=(\sqrt{2}a+ut)\frac{d}{dt}(a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}\:)^\frac{1}{2}+(\sqrt{a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}}\:)[\frac{d}{dt}(\sqrt{2}a)+\frac{d}{dt}(ut)]}$

$\sf{=(\sqrt{2}a+ut).\frac{1}{2}(a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}\:)^\frac{-1}{2}+(\sqrt{a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}}\:)[0+u\frac{d}{dt}(t)]}$

$\sf{=\frac{\frac{1}{2}(\sqrt{2}a+ut)}{(a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}\:)^\frac{1}{2}}+(\sqrt{a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}}\:)[0+u(1)]}$

$\sf{=\frac{\frac{1}{2}(\sqrt{2}a+ut)}{(a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}\:)^\frac{1}{2}}+(a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}\:)^{\frac{1}{2}}[u]}$

$\sf{=\frac{\frac{1}{2}(\sqrt{2}a+ut)+(a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}\:)[u]}{(a^{2}-(\frac{a}{\sqrt{2}}+ut)^{2}\:)^{\frac{1}{2}}}}$