what is the derivative of the integral from 2 to 1/X of sin^4 t dt

Question

anonymous · Answer

$$f(x)=\int_2^x\sin^4(t)dt$$
$$f'(x)=\sin^4(x)$$ because the derivative of the integral is the intergrand
thefore
$$f(\frac{1}{x})=\int_0^{\frac{1}{x}}\sin^4(t)dt$$
$$f'(\frac{1}{x})=\frac{-1}{x^2}\sin^4(\frac{1}{x})$$ by the chain rule

anonymous · Answer

typo as i changed the lower limit of integration but it does not change the answer

anonymous · Answer

good morning all ( at least morning here)

anonymous · Answer

Use Leibniz integral rule http://en.wikipedia.org/wiki/Leibniz_integral_rule to get your answer and Good evening friends (here)

anonymous · Answer

@ellex11  you got this?

anonymous · Answer

@satellite73 I think you lost the OP when you posted the answer without engaging any discussion what-so-ever.  I don't believe this is a race to see how fast WE can answer the question, rather how WELL we can help the OP.  just a thought.

anonymous · Answer

It's actually very easy applying Leibnitz's theorem,

$$F(x) = \large\int \limits_2 ^ {\frac{1}{x}} \sin ^4 t \; dt $$
Applying Leibnitz's theorem, we get, $$ F'(x) = \sin^4 \left(\frac{1}{x} \right) \cdot \frac d{dx} \frac{1}{x} -  \sin^4 \left(2 \right)  \cdot \frac d{dx} (2) $$

$$ = -\frac 1 {x^2} \sin^4 \left(\frac{1}{x} \right)$$

anonymous · Answer

true, but kind of weird strange since you have a constant

anonymous · Answer

Sat, for a monovariant function $g$:

$$ {d\over dx}\, \int_{f_1(x)}^{f_2(x)} g(t) \,dt = g(f_2(x)) {f_2'(x)} -  g(f_1(x)) {f_1'(x)} $$

can't we assume $ f_1(x) = 2 $?

anonymous · Answer

yes yes, of course it works.

anonymous · Answer

Aha!, thanks :)