M myngo9191 Mar 2010 15 0 May 9, 2010 #1 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 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?

AllanCuz Apr 2010 384 153 Canada May 9, 2010 #2 myngo9191 said: 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? Click to expand... 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?

myngo9191 said: 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? Click to expand... 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?

M myngo9191 Mar 2010 15 0 May 9, 2010 #3 AllanCuz said: 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? Click to expand... Is this what I should have? \(\displaystyle (4lnx^2sinlnx^2)^\frac{1}{3} \frac{2x}{x^2} \Bigg|_{-1}^{lnx^2} \)

AllanCuz said: 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? Click to expand... Is this what I should have? \(\displaystyle (4lnx^2sinlnx^2)^\frac{1}{3} \frac{2x}{x^2} \Bigg|_{-1}^{lnx^2} \)

AllanCuz Apr 2010 384 153 Canada May 9, 2010 #4 myngo9191 said: Is this what I should have? \(\displaystyle (4lnx^2sinlnx^2)^\frac{1}{3} \frac{2x}{x^2} \Bigg|_{-1}^{lnx^2} \) Click to expand... 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! Reactions: myngo9191

myngo9191 said: Is this what I should have? \(\displaystyle (4lnx^2sinlnx^2)^\frac{1}{3} \frac{2x}{x^2} \Bigg|_{-1}^{lnx^2} \) Click to expand... 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!

M myngo9191 Mar 2010 15 0 May 9, 2010 #5 AllanCuz said: 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! Click to expand... 0o0h!!! Thank you s0o0 much!!! (Yes)

AllanCuz said: 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! Click to expand... 0o0h!!! Thank you s0o0 much!!! (Yes)