Question

: What is the first derivative of:

\[y = 4 + \int_{0}^{x}(5t-4y(t))dt\]

Answer 1

\[\frac{d}{dx}\int_{0}^{x}(5t-4y(t))dt\]
\[\frac{d}{dx}\int_{0}^{x}5t\ dt-\int_{0}^{x}4y(t)\ dt\]
\[\frac{d}{dx}5\int_{0}^{x}t\ dt-4\int_{0}^{x}y(t)\ dt\]
\[\frac{d}{dx}5(\frac{x^2}{2}-\frac{0^2}{2})-4(Y(x)-Y(0))\]
\[\frac{d}{dx}\frac{5x^2}{2}-\frac{d}{dx}4Y(x)-\frac{d}{dx}4Y(0)\]
\[\frac{dx}{dx}5x^2-\frac{dx}{dx}4y(x)-\frac{dx}{dx}0\]
\[5x^2-4y(x)\]
maybe

Answer 2

i forgot to undo the ^2 on the 5x^2 lol

Answer 3

By the fundamental theorem of calculus:

dy/dx = the integrand = 5x - 4y(x)

(the 4 out front is irrelevent, it's derivative is zero.)

Answer 4

That is to say
\[\frac{d\ }{dx} \int\limits_a^x f(t) dt \ \ \ = \ \ \ f(x)\]