$$f(x,y)=\int_{e^{4y}}^{\ln^3(x)}{\frac{\sin(t)}{t}\,dt}$$
Whats the derivative $\frac{d f}{d t}$, if:
$$x(t)=\cos(2+6t).4t^2$$ $$y(t)=\ln(2r+7e^{5t})$$
Really not much to say about this problem since I never saw it before.
Ps: Edition has been made but the original question used $\partial f/\partial t$ (partial derivative instead of $d$). Also, if is worth adding, the first part of the question asked to calculate $\partial f/\partial x$ and $\partial f/\partial y$, but did not caused me trouble. I would really appreciate if someone could solve this.
Use the Liebniz integral rule to get the result. That is, $$ \frac{d}{dt}\int_{a(t)}^{b(t)}F(w,t)dw = \int_{a(t)}^{b(t)}\frac{\partial F}{\partial t}(w,t)dw + F(b(t),t)\frac{d b}{dt}(t)-F(a(t),t)\frac{d a}{dt} $$
In this case you have $$ b(t)=\log^3(x(t))=4t^2 \cos(2+6t) $$ $$ a(t) = \exp(4y(t))=\exp(4\log(2r+7\exp(5(t)))) $$ and $$ F(w,t) = \frac{\sin(w)}{w} $$
Notice the last is independent of a time variable $t$, any variable in the integrand is participating in the integration and is in fact a different $t$ than what you are differentiating.