Step 1Given expressiony=eexStep 2Using the chain rule of derivativesIf y=f(u), where u is a function of x.Then dydx=dydu⋅dudxStep 3Apply the chain rule of derivative in the given expressionLet u=exThen y=euNowdydx=dydu⋅dudxy′=d(eu)du⋅dudxy′=(eu)⋅dudx....(1)Step 4Putting the value of u equation (1)y′=(eex)⋅d(ex)dxy′=(eex)⋅exy′=ex(eex)...(2)Step 5For the second derivativeUsing product rule of derivativeIf y=u(x)v(x) then dydx=u(x)v′(x)+v(x)u′(x)Step 6Apply the product rule of derivatives in equation (2)y″=ex⋅d(eex)dx+(eex)⋅d(ex)dxy″=ex⋅(ex(eex))+(eex)⋅exy″=ex(eex)(ex+1)

Best solutions to - "Find y"

Chesley

Answered question

2021-11-08

Find y

Answer & Explanation

Gennenzip

Skilled2021-11-09Added 96 answers

Step 1
Given expression

y = e^{e^{x}}

Step 2
Using the chain rule of derivatives
If y=f(u), where u is a function of x.
Then

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}

Step 3
Apply the chain rule of derivative in the given expression
Let

u = e^{x}

Then

y = e^{u}

Now

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}

y^{'} = \frac{d (e^{u})}{d u} \cdot \frac{d u}{d x}

y^{'} = (e^{u}) \cdot \frac{d u}{d x}

....(1)
Step 4
Putting the value of u equation (1)

y^{'} = (e^{e^{x}}) \cdot \frac{d (e^{x})}{d x}

y^{'} = (e^{e^{x}}) \cdot e^{x}

y^{'} = e^{x} (e^{e^{x}})

...(2)
Step 5
For the second derivative
Using product rule of derivative
If y=u(x)v(x) then

\frac{d y}{d x} = u (x) v^{'} (x) + v (x) u^{'} (x)

Step 6
Apply the product rule of derivatives in equation (2)

y^{″} = e^{x} \cdot \frac{d (e^{e^{x}})}{d x} + (e^{e^{x}}) \cdot \frac{d (e^{x})}{d x}

y^{″} = e^{x} \cdot (e^{x} (e^{e^{x}})) + (e^{e^{x}}) \cdot e^{x}

y^{″} = e^{x} (e^{e^{x}}) (e^{x} + 1)

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