Question

53-56 Find \(y'\) and \(y''\).

56. \(y = {e^{{e^x}}}\)

Short answer

The value of \(y'\) is \({e^{{e^x} + x}}\).

The value of \(y''\) is \({e^{{e^x} + x}}\left( {{e^x} + 1} \right)\).

Step-by-step solution

Step 1:Find the value of \(y'\) 

Find the value of \(y'\) for the equation\(y = {e^{{e^x}}}\).

\(\begin{aligned}\frac{{{\rm{d}}y}}{{{\rm{d}}x}} &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{e^{{e^x}}}} \right)\\ &= {e^{{e^x}}}\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{e^x}} \right)\\ &= {e^{{e^x}}} \cdot {e^x}\\& = {e^{{e^x} + x}}\end{aligned}\)

So, the value of \(y'\) is \({e^{{e^x} + x}}\).

Find the value of \(y''\)

Find the value of \(y''\) from the equation \(y' = {e^{{e^x} + x}}\) using the quotient rule.

\(\begin{aligned}y'' &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{e^{{e^x} + x}}} \right)\\ &= \left( {{e^{{e^x} + x}}} \right) \cdot \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{e^x} + x} \right)\\ &= {e^{{e^x} + x}}\left( {{e^x} + 1} \right)\end{aligned}\)

So, the value of \(y''\) is \({e^{{e^x} + x}}\left( {{e^x} + 1} \right)\).