Question

Question #81d94

Answer 1

`y^'=x^(x^2+1)(2ln(x)+1)`

Explanation:

First, use the rule that `(a^b)^c=a^((bc)` so `y=x^((x*x))=x^(x^2)`. We have:

`y=x^(x^2)`

Take the natural logarithm of both sides:

`ln(y)=ln(x^(x^2))`

Simplifying this using `log(a^b)=blog(a)`:

`ln(y)=x^2ln(x)`

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=fg`, then `y^'=f^'g+fg^')`.

We get:

`1/y*y^'=(x^2)^'ln(x)+x^2(ln(x))^'`

The derivative of `x^2` is `2x` and of `ln(x)` is `1/x`:

`1/y*y^'=2xln(x)+x^2(1/x)`

`1/y*y^'=x(2ln(x)+1)`

Now, solve for `y^'` by multiplying both sides of the equation by `y`.

`y^'=yx(2ln(x)+1)`

Since `y=x^(x^2)`:

`y^'=x^(x^2)x(2ln(x)+1)`

Simplify `x^(x^2)x` using the rule `a^ba^c=a^(b+c)`:

`y^'=x^(x^2+1)(2ln(x)+1)`