Answer:
[tex]y'=\frac{7t^2-10t-21}{(7t-5)^2}[/tex]
Step-by-step explanation:
We have been given a function [tex]y=\frac{t^2+3}{7t-5}[/tex]. We are asked to find the derivative of the given function.
We will use quotient rule of derivatives to find derivative of our given function.
[tex]\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}[/tex]
[tex]y'=\frac{\frac{d}{dt}(t^2+3)\cdot (7t-5)-\frac{d}{dt}(7t-5)\cdot(t^2+3)}{(7t-5)^2}[/tex]
[tex]y'=\frac{(2t)\cdot (7t-5)-7\cdot(t^2+3)}{(7t-5)^2}[/tex]
[tex]y'=\frac{14t^2-10t-7t^2-21}{(7t-5)^2}[/tex]
[tex]y'=\frac{7t^2-10t-21}{(7t-5)^2}[/tex]
Therefore, our required derivative would be [tex]y'=\frac{7t^2-10t-21}{(7t-5)^2}[/tex].
