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

Question

Emily Ammerman · Accepted Answer

#-(12z+6)/(z^2+z-1)^2# #"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#

Aurora Alvarado · Answer

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#

Adam Adkins · Answer

undefined To find the derivative of $6(z^2+z-1)^{-1}$, you can use the chain rule. The derivative is:

$$ \frac{d}{dz} \left[6(z^2+z-1)^{-1}\right] = -6(z^2+z-1)^{-2} \cdot \frac{d}{dz}(z^2+z-1) $$

Applying the chain rule to $\frac{d}{dz}(z^2+z-1)$:

$$ \frac{d}{dz}(z^2+z-1) = 2z + 1 $$

Substituting back into the original expression:

$$ \frac{d}{dz} \left[6(z^2+z-1)^{-1}\right] = -6(z^2+z-1)^{-2} \cdot (2z + 1) $$

So, the derivative is:

$$ -6(2z + 1)(z^2+z-1)^{-2} $$