# Question 81d94

Aug 27, 2016

${y}^{'} = {x}^{{x}^{2} + 1} \left(2 \ln \left(x\right) + 1\right)$

#### Explanation:

First, use the rule that (a^b)^c=a^((bc) so $y = {x}^{\left(x \cdot x\right)} = {x}^{{x}^{2}}$. We have:

$y = {x}^{{x}^{2}}$

Take the natural logarithm of both sides:

$\ln \left(y\right) = \ln \left({x}^{{x}^{2}}\right)$

Simplifying this using $\log \left({a}^{b}\right) = b \log \left(a\right)$:

$\ln \left(y\right) = {x}^{2} \ln \left(x\right)$

Differentiate both sides. On the left side, the chain rule will kick into effect, since this is a function of $y$ and not $x$. On the right hand side, we will use the product rule (if $y = f g$, then y^'=f^'g+fg^')#.

We get:

$\frac{1}{y} \cdot {y}^{'} = {\left({x}^{2}\right)}^{'} \ln \left(x\right) + {x}^{2} {\left(\ln \left(x\right)\right)}^{'}$

The derivative of ${x}^{2}$ is $2 x$ and of $\ln \left(x\right)$ is $\frac{1}{x}$:

$\frac{1}{y} \cdot {y}^{'} = 2 x \ln \left(x\right) + {x}^{2} \left(\frac{1}{x}\right)$

$\frac{1}{y} \cdot {y}^{'} = x \left(2 \ln \left(x\right) + 1\right)$

Now, solve for ${y}^{'}$ by multiplying both sides of the equation by $y$.

${y}^{'} = y x \left(2 \ln \left(x\right) + 1\right)$

Since $y = {x}^{{x}^{2}}$:

${y}^{'} = {x}^{{x}^{2}} x \left(2 \ln \left(x\right) + 1\right)$

Simplify ${x}^{{x}^{2}} x$ using the rule ${a}^{b} {a}^{c} = {a}^{b + c}$:

${y}^{'} = {x}^{{x}^{2} + 1} \left(2 \ln \left(x\right) + 1\right)$