Mathematics 19 Online
OpenStudy (amistre64):

: What is the first derivative of: $y = 4 + \int_{0}^{x}(5t-4y(t))dt$

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (jamesj):

By the fundamental theorem of calculus: dy/dx = the integrand = 5x - 4y(x) (the 4 out front is irrelevent, it's derivative is zero.)

OpenStudy (jamesj):

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