This is the equation: $\displaystyle \frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt $ and this is what I have: $\displaystyle (4xsinx)^\frac{1}{3} \Bigg|_{-1}^{lnx^2}$ Now where do I go from here?
Follow Math Help Forum on Facebook and Google+
Originally Posted by myngo9191 This is the equation: $\displaystyle \frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt $ and this is what I have: $\displaystyle (4xsinx)^\frac{1}{3} \Bigg|_{-1}^{lnx^2}$ Now where do I go from here? This follows from the fundemental theorem of calculus and what have there is not correct. $\displaystyle \frac{d}{dx} \int_{a}^{g(x)} f(t)dt = f[g(x)]g`(x) $ Can you compute knowing the above?
Originally Posted by AllanCuz This follows from the fundemental theorem of calculus and what have there is not correct. $\displaystyle \frac{d}{dx} \int_{a}^{g(x)} f(t)dt = f[g(x)]g`(x) $ Can you compute knowing the above? Is this what I should have? $\displaystyle (4lnx^2sinlnx^2)^\frac{1}{3} \frac{2x}{x^2} \Bigg|_{-1}^{lnx^2} $
Originally Posted by myngo9191 Is this what I should have? $\displaystyle (4lnx^2sinlnx^2)^\frac{1}{3} \frac{2x}{x^2} \Bigg|_{-1}^{lnx^2} $ From above, $\displaystyle \frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt = [4(lnx^2)sin(lnx^2)]^\frac{1}{3} * \frac{2}{x} $ We do not evaluate here which is what you have in your post. We are finished, this is the answer!
Originally Posted by AllanCuz From above, $\displaystyle \frac{d}{dx} \int_{-1}^{lnx^2} (4tsint)^\frac{1}{3} dt = [4(lnx^2)sin(lnx^2)]^\frac{1}{3} * \frac{2}{x} $ We do not evaluate here which is what you have in your post. We are finished, this is the answer! 0o0h!!! Thank you s0o0 much!!!
View Tag Cloud