How do you differentiate $y = 2^{3^{x^{2}}}$ $y=2^(3^(x^2))$ ?

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How do you differentiate $y = 2^{3^{x^{2}}}$ $y=2^(3^(x^2))$ ?

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Answer 1 · 2017-01-30T16:26:43

$\frac{d y}{d x} = 2^{3^{x^{2}}} \times 3^{x^{2}} ln 18 \times x$ $(dy)/(dx)=2^(3^(x^2))xx3^(x^2)ln18xx x$

Explanation:

As $y = 2^{3^{x^{2}}}$ $y=2^(3^(x^2)$ , we have

$ln y = 3^{x^{2}} ln 2$ $lny=3^(x^2)ln2$ or $3^{x^{2}} = \frac{ln y}{ln 2}$ $3^(x^2)=lny/ln2$

i.e. $x^{2} ln 3 = ln (\frac{ln y}{ln 2})$ $x^2ln3=ln(lny/ln2)$

and hence differentiating we get

$2 ln 3 \times x = \frac{1}{(\frac{ln y}{ln 2})} \times \frac{1}{ln 2} \times \frac{1}{y} \frac{d y}{d x}$ $2ln3xx x=1/((lny/ln2))xx1/ln2xx1/y(dy)/(dx)$

or $2 ln 3 \times x = \frac{1}{ln y} \times \frac{1}{y} \frac{d y}{d x}$ $2ln3xx x=1/lnyxx1/y(dy)/(dx)$

and $\frac{d y}{d x} = y \times ln y \times 2 ln 3 \times x$ $(dy)/(dx)=yxxlnyxx2ln3xx x$

= $2^{3^{x^{2}}} \times 3^{x^{2}} ln 2 \times 2 ln 3 \times x$ $2^(3^(x^2))xx3^(x^2)ln2xx2ln3xx x$

= $2^{3^{x^{2}}} \times 3^{x^{2}} ln 18 \times x$ $2^(3^(x^2))xx3^(x^2)ln18xx x$

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How do you differentiate $y = 2^{3^{x^{2}}}$ $y=2^(3^(x^2))$ ?

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