How would I find the derivatives of these functions?
1. $\displaystyle f(x) = \int_{x}^{1} \sqrt{t+sin(t)} dt $
2. $\displaystyle y = \int_{\sqrt{x}}^{x} \frac{cos\theta}{\theta} d\theta $
So is this true then?
1. $\displaystyle f(x) = \int_{x}^{1} \sqrt{t+sin(t)} dt $
Let $\displaystyle g(t)=\sqrt{t+sin(t)$ so that $\displaystyle f(x) = \int_{x}^{1} g(t) dt$
Then $\displaystyle D\left( f(x) \right) = D\left( \int_{x}^{1} g(t) dt \right) = D\left( G(1) - G(x) \right) = D\left( G(1) \right) - D\left( G(x) \right) = D\left( -G(x) \right) = -g(x) $
2. $\displaystyle y = \int_{\sqrt{x}}^{x} \frac{cos\theta}{\theta} d\theta $
Let $\displaystyle y = f(x)$ and $\displaystyle g(\theta)=\sqrt{\theta+sin(\theta)$ so that $\displaystyle f(x) = \int_{x}^{1} g(\theta) d\theta$
Then $\displaystyle D\left( f(x) \right) = D\left( \int_{x}^{1} g(\theta) d\theta \right) = D\left( G(x) - G(\sqrt{x}) \right) = g(x) - g(\sqrt{x})$
Seems correct, but it seems too simple. I suppose these problems could be just to illustrate how integration and differentiation are inverse processes.
Do not make this hard.
$\displaystyle D_x \left( {\int_0^x {f(t)dt} } \right)=f(x)$
Because $\displaystyle {\int_x^0 {f(t)dt}=-{\int_0^x {f(t)dt}$ we have $\displaystyle D_x \left( {\int_x^0 {f(t)dt} } \right)=-f(x)$
$\displaystyle D_x \left( {\int_{g(x)}^{h(x)} {f(t)dt} } \right) = f \circ h(x)h'(x) - f \circ g(x)g'(x)$
Right. Thanks. "KISS = Keep It Simple, Stupid" I'll try to remember that.
I'm going with these answers then:
1. $\displaystyle D(f(x)) = -\sqrt{x+sin(x)} $
2. $\displaystyle D(y) = \frac{cos(x)-\sqrt{x} \, cos\sqrt{x}}{x} $
(unless someone says "no, don't" - anyone?)
Since Plato edited his response to include the above, it appears at least one of my answers is wrong.
2. $\displaystyle y = \int_{\sqrt{x}}^{x} \frac{cos\theta}{\theta} d\theta $
Let $\displaystyle y = f(x)$ and $\displaystyle t(\theta)=\frac{cos\theta}{\theta}$ so that $\displaystyle f(x) = \int_{\sqrt{x}}^{x} t(\theta) d\theta$
Also, if $\displaystyle g(x) = \sqrt{x}$ and $\displaystyle h(x) = x$ then this would conform with Plato's formula and the derivatives would be $\displaystyle h'(x) = 1, g'(x)=\frac{1}{2\sqrt{x}}$.
The integral would NOT be what I posted earlier: $\displaystyle D(y) = \frac{cos(x)-\sqrt{x} \, cos\sqrt{x}}{x} $.
Instead, this is what I'm thinking:
$\displaystyle D\left( f(x) \right) = D\left( \int_{g(x)}^{h(x)} t(\theta) d\theta \right) = t(x)*1 - t(\sqrt{x})*\frac{1}{2\sqrt{x}} $
$\displaystyle = \frac{cos(x)}{x}*1 - \frac{cos(\sqrt{x})}{\sqrt{x}}*\frac{1}{2\sqrt{x}} $
$\displaystyle = \frac{2cos(x)-cos(\sqrt{x})}{2x}$
Because the derivative of $\displaystyle x$ is 1, I'm sticking with my original results for the first problem, which was $\displaystyle D(f(x)) = -\sqrt{x+sin(x)} $
Once again, can anyone confirm or deny these answers? Thanks.
More generally, "Liebniz's rule":
$\displaystyle \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= \frac{d\beta(x)}{dx}f(x,\beta(x))- \frac{d\alpha(x)}{dx}f(x, \alpha(x)) + \int_{\alpha(x)}^{\beta(x)}\frac{\partial f}{\partial x} dt$