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Answer 1 · 2017-04-09T19:14:43

I'm assuming your question is:

What's $\frac{d y}{d x}$ $dy/dx$ when $y = {(x + 1)}^{x^{2}}$ $y=(x+1)^(x^2)$ ?

$y = {(x + 1)}^{x^{2}}$ $y=(x+1)^(x^2)$

Take the natural logarithm of both sides:

$ln (y) = ln ({(x + 1)}^{x^{2}})$ $ln(y)=ln((x+1)^(x^2))$

$ln (y) = x^{2} ln (x + 1)$ $ln(y)=x^2ln(x+1)$

Differentiate both sides.

$\frac{1}{y} \frac{d y}{d x} = 2 x ln (x + 1) + \frac{x^{2}}{x + 1}$ $1/ydy/dx=2xln(x+1)+x^2/(x+1)$

$\frac{d y}{d x} = y (2 x ln (x + 1) + \frac{x^{2}}{x + 1})$ $dy/dx=y(2xln(x+1)+x^2/(x+1))$

$\frac{d y}{d x} = {(x + 1)}^{x^{2}} (2 x ln (x + 1) + \frac{x^{2}}{x + 1})$ $dy/dx=(x+1)^(x^2)(2xln(x+1)+x^2/(x+1))$