Answer:
f'(y) = [tex]9sec^2(y)-\frac{1}{8\sqrt{y} } - \frac{5y^5\sqrt{y} - 50y^4\sqrt{y} }{y^{10}}[/tex] or any equivalent form
Step-by-step explanation:
rewriting the equation to contain only exponents no roots
[tex]f(y)= 9tan(y)-\frac{1}{4} y^\frac{1}{2} -\frac{10y^\frac{1}{2}}{y^5}[/tex]
using power rule, quotient rule, and trig derivatives
f'(y) = [tex]9sec^2(y)-\frac{1}{8\sqrt{y} } - \frac{\frac{5y}{\sqrt{y}}(y^5) - 5y^4(10y^\frac{1}{2}) }{y^{10}}[/tex]
f'(y) = [tex]9sec^2(y)-\frac{1}{8\sqrt{y} } - \frac{5y^5\sqrt{y} - 50y^4\sqrt{y} }{y^{10}}[/tex]
