I got problem with the derivative of y=a^2x (1 Viewer)

[icycloud]

It's just the chain rule.
Take y = x^x

ln(y) = xln(x)

You differentiate with respect to x on both sides, so:

on the LHS, let z = ln(y), then dz/dx = dz/dy * dy/dx
dz/dx = 1/y * dy/dx

so the derivative of LHS is 1/y *dy/dx

RHS you just differentiate normally using the product rule, yielding: ln(x) + 1

And so 1/y * dy/dx = ln(x) + 1
dy/dx = y(ln(x)+1)

Subbing in y = x^x, we have:

dy/dx = x^x (ln(x)+1)

 

[abcd9146]

my maths teacher just said totday that logarithmic differentiation was a concept you learn in first year university... so its probably true, we dont really need it , but its still interesting to learn about this kind of stuff

 

[Riviet]

Implicit differentiation is not examinable in the 2/3 unit HSC, but 4 unit students of course need to know it.