Derivative by First Principal for Reciprocal Square Root

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lemonadatv

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japojo

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DianaTsisika-oucf

It may be a good exercise to prove it from first principles .For this and similar examples it is much more elegant to use the product rule.
Set y[x]= √(x+2) . Then y[x]^2 *(x+2) =1 .By the product rule we have (y[x]^2)' *(x+2) +y[x]^2 *1 = 0 .Again by the product rule
we have (y[x]^2) ' = 2*y ' [x] *y[x] . Hence, by solving for y ' [x] we get y ' [x]= -1/2 (x+2)^(-3/2) .
This works for any rational power of a*x+b .

renesperb

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allanchichi

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JOKU-

Sir can u give me the solution of find the derivative of cos(ax^2+bx+c)from first principle

mithunsaha

What if there is a bigger number on top

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jaysonbalanday

Excuse me Sir please give a solution for derivative of x root under x+a by root under x square+ a square

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GamingwithVishnuYT

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MalikHarisAbdullahAwan

Sir aap class11 ka math root cosx and root secx wala question dijye ga

deepalibauri

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