How do you find the derivative of #f(z)= (z^2+1)/(sqrt z)#?

Question

Cameron Alexander · Accepted Answer

#f'(z)=(3z^2-1)/(2sqrt(z^3))# #"differentiate using the "color(blue)"quotient rule"# #"given "f(z)=(g(z))/(h(z))" then"# #f'(z)=(h(z)g'(z)-g(z)h'(z))/(h(z))^2larrcolor(blue)"quotient rule"# #g(z)=z^2+1rArrg'(z)=2z# #h(z)=z^(1/2)rArrh'(z)=1/2z^(-1/2)# #rArrf'(z)=(2z^(3/2)-1/2z^(-1/2)(z^2+1))/z# #color(white)(rArrf'(z))=(1/2z^(-1/2)(4z^2-z^2-1))/z# #color(white)(rArrf'(z))=(3z^2-1)/(2z^(3/2))=(3z^2-1)/(2sqrt(z^3)#

Ezra Adams · Answer

undefined To find the derivative of $ f(z) = \frac{z^2 + 1}{\sqrt{z}} $, we can use the quotient rule, which states that if $ u $ and $ v $ are differentiable functions of $ z $, then the derivative of $ \frac{u}{v} $ with respect to $ z $ is given by $ \frac{u'v - uv'}{v^2} $, where $ u' $ and $ v' $ denote the derivatives of $ u $ and $ v $ with respect to $ z $, respectively. Applying this rule to $ f(z) $, we get:

$$ f'(z) = \frac{(2z)(\sqrt{z}) - (z^2 + 1)\left(\frac{1}{2\sqrt{z}}\right)}{(\sqrt{z})^2} $$

Simplify the expression:

$$ f'(z) = \frac{2z\sqrt{z} - \frac{z^2 + 1}{2\sqrt{z}}}{z} $$

$$ f'(z) = \frac{2z^{\frac{3}{2}} - \frac{z^2 + 1}{2z^{\frac{1}{2}}}}{z} $$

$$ f'(z) = \frac{2z^{\frac{5}{2}} - (z^2 + 1)z^{-\frac{1}{2}}}{2z} $$

$$ f'(z) = \frac{2z^{\frac{5}{2}} - z^{\frac{5}{2}} - z^{-\frac{1}{2}}}{2z} $$

$$ f'(z) = \frac{z^{\frac{5}{2}} - z^{-\frac{1}{2}}}{2z} $$

$$ \boxed{f'(z) = \frac{z^{\frac{5}{2}} - 1}{2z}} $$