# `y = sqrt(x + sqrt(x + sqrt(x)))` Find the derivative of the function.

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### 3 Answers

`y=sqrt(x+sqrt(x+sqrt(x)))`

`y'=(1/(2sqrt(x+sqrt(x+sqrt(x))))) *d/dx (x+sqrt(x+sqrt(x)))`

`y'=(1/(2sqrt(x+sqrt(x+sqrt(x))))) *(1+d/dxsqrt(x+sqrt(x)))`

`y'=(1/(2sqrt(x+sqrt(x+sqrt(x))))) *(1+(1/(2sqrt(x+sqrt(x)))*d/dx (x+sqrt(x))))`

`y'=(1/(2sqrt(x+sqrt(x+sqrt(x))))) *(1+(1/(2sqrt(x+sqrt(x))))*(1+d/dxsqrt(x)))`

`y'=(1/(2sqrt(x+sqrt(x+sqrt(x))))) *(1+(1/(2sqrt(x+sqrt(x)))) *(1+1/(2sqrt(x))))`

`y'=(1/(2sqrt(x+sqrt(x+sqrt(x))))) *(1+(1/(2sqrt(x+sqrt(x)))) *((2sqrt(x)+1)/(2sqrt(x))))`

`y'=(1/(2sqrt(x+sqrt(x+sqrt(x))))) *(1+(2sqrt(x)+1)/(4sqrt(x)sqrt(x+sqrt(x))))`

`y'=(1/(2sqrt(x+sqrt(x+sqrt(x))))) *((4sqrt(x)sqrt(x+sqrt(x))+2sqrt(x)+1)/(4sqrt(x)sqrt(x+sqrt(x))))`

`y'=(4sqrt(x)sqrt(x+sqrt(x))+2sqrt(x)+1)/(8sqrt(x)sqrt(x+sqrt(x))sqrt(x+sqrt(x+sqrt(x))))`

Note:- 1) If y = sqrt(x) ; then dy/dx = 1/2sqrt(x)

Now,

y = sqrt[x + sqrt{(x) + sqrt(x)}]

Apply, the chain rule i.e ,

if y = u*v ; where u & v botyh are functions of 'x' ; then

dy/dx = u*(dv/dx) + v*(du/dx)

Also, apply the rule of outer to inner when there are subfunctions inside a function.

Please find the solution in the attachment.

### Hide Replies ▲

Answer posted is wrong.

`y=sqrt(x+sqrt(x+sqrt(x)))`

`y'=(1/(2sqrt(x+sqrt(x+sqrt(x))))).(1+(1/(2sqrt(x+sqrt(x))))).(1+(1/(2sqrt(x))))`

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