Understanding the Derivative of 2^(x^2)

[aguycalledwil]

I'm trying to differentiate 2^(x^2), but I'm getting a factor of two out and can't figure out why. I approached the question as follows..

y=2^(x^2) , so y=(2^x)^x
u=2^x y=u^x

du/dx = (2^x)ln2
dy/du = xu^(x-1)
= x(2^x)^(x-1)
= x(2)^((x^2)-x)

So dy/dx =
[x(2)^((x^2)-x)]*[(2^x)Ln2]

However, on the mark scheme it says when x=2, the gradient should be 64ln2. Using my derivative, at x=2 the gradient comes out at 32ln2. Can anyone help me find where I've gone wrong? Much appreciated!

 

[hamsterman]

$2^{x^2}$ is a composition of functions $f(x) = 2^x$ and $g(x) = x^2$. You know that $f'(x) = 2^x \ln 2$ and $g'(x) = 2x$

There is a formula for derivative of composite functions. $(f \circ g)'(x) = f'(g(x))g'(x)$. This is just another form of the chain rule. After blindly pasting the functions we already have, we get $2x \cdot 2^{x^2} \ln 2$

I can't see what error you made as those formulas are not very readable. Try using latex.

 

[Curious3141]

I'm trying to differentiate 2^(x^2), but I'm getting a factor of two out and can't figure out why. I approached the question as follows..

y=2^(x^2) , so y=(2^x)^x
u=2^x y=u^x

du/dx = (2^x)ln2
dy/du = xu^(x-1)

That's wrong right there. The rule $\frac{d}{dx}x^n = nx^{n-1}$ ONLY applies when n is a constant. If n is a variable like x or a function of x, the rule simply does not work.

Your best bet here is to use Chain Rule as hamsterman mentioned.

 

[hotvette]

Another approach is to take the (natural) log of both sides of y = 2^x2 and differentiate implicitly.

 

[aguycalledwil]

Thanks guys, got it!