(1)find dy/dx of y if y=tan^(-1)1/x^2

Question

(2) if y=3z^2.z^2 find dy/dz
(3) if y=x^x show that dy/dx=x|n(x+1).. thanks

Answer 1

just good old chain rule stuff

If y = arctan(u),
y' = 1/(1+u^2) u'
= 1/(1+(1/x^2)^2) (-2/x^3)
= -2x/(1+x^4)

y = 3z^2 * z^2 = 3z^4
y' = 12z^3
Seems too easy. Maybe I misread the problem

y = x^x
lny = x lnx
1/y y' = lnx + 1
y' = x^x (1+lnx)

I find it interesting that
d/dx u^n = n u^(n-1) u'
d/dx a^v = lna a^v

d/dx u^v = v u^(v-1) u' + lnu u^v v'
= u^v (v/u u' + lnu v')
You have u=v=x, so
y' = x^x (1 + lnx)
It's just a combination of the exponent and power rules.

Answer 2

2X+4x =6x

Answer 3

i can do that
2x+4x=6x
6x=6x add -6x to both side
6x-6x=6x-6x
0=0 which automatically mean that all real no are the solution

Answer 4

that is correct. No matter what value you pick for x, 2x+4x=6x.