Question

How do you find the derivative of `6(z^2+z-1)^-1`?

Answer 1

`-(12z+6)/(z^2+z-1)^2`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given "y=f(g(x)" then"`

`dy/dx=f'(g(x))xxg'(x)larr" chain rule"`

`d/dz(6(z^2+z-1)^-1)`

`=-6(z^2+z-1)^-2xxd/dz(z^2+z-1)`

`=(-6(2z+1))/(z^2+z-1)^2=-(12z+6)/(z^2+z-1)^2`

Answer 2

Recall the power rule: `d/dxx^n=nx^(n-1)`. When we have a function to a power, we still use the power rule to differentiate it, but we do so as well as using the chain rule.

Combining the chain rule with the power rule for some function `u` gives us: `d/dxu^n=n u^(n-1)(du)/dx`

Thus:

`d/(dz)6(z^2+z-1)^-1=6(-1(z^2+z-1)^-2)d/(dz)(z^2+z-1)`

And we can use the power rule to find the derivative of `z^2+z-1`:

`d/(dz)6(z^2+z-1)^-1=-6(z^2+z-1)^-2(2z+1)`

`=color(blue)((-6(2z+1))/(z^2+z-1)^2`