Please refer to my attachment - I'm not sure why this method does not work... Thank you!
E eutas1 Junior Member Joined Apr 8, 2021 Messages 165 Apr 13, 2021 #1 Please refer to my attachment - I'm not sure why this method does not work... Thank you! Attachments 777.JPG 61.2 KB · Views: 3
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Apr 13, 2021 #2 [MATH]\frac{d}{dx}(e^{2x} \cdot e^{\sin(4x)})[/MATH] requires use of the product rule [MATH]e^{2x} \cdot e^{\sin(4x)} \cdot 4\cos(4x) + e^{\sin(4x)} \cdot 2e^{2x} = 2e^{2x} \cdot e^{\sin(4x)}[2\cos(4x)+1][/MATH]
[MATH]\frac{d}{dx}(e^{2x} \cdot e^{\sin(4x)})[/MATH] requires use of the product rule [MATH]e^{2x} \cdot e^{\sin(4x)} \cdot 4\cos(4x) + e^{\sin(4x)} \cdot 2e^{2x} = 2e^{2x} \cdot e^{\sin(4x)}[2\cos(4x)+1][/MATH]
E eutas1 Junior Member Joined Apr 8, 2021 Messages 165 Apr 13, 2021 #3 skeeter said: [MATH]\frac{d}{dx}(e^{2x} \cdot e^{\sin(4x)})[/MATH] requires use of the product rule [MATH]e^{2x} \cdot e^{\sin(4x)} \cdot 4\cos(4x) + e^{\sin(4x)} \cdot 2e^{2x} = 2e^{2x} \cdot e^{\sin(4x)}[2\cos(4x)+1][/MATH] Click to expand... Ah I see! However, I still don't get exactly the same answer - please refer to the attachment. Attachments 111.jpg 32 KB · Views: 4
skeeter said: [MATH]\frac{d}{dx}(e^{2x} \cdot e^{\sin(4x)})[/MATH] requires use of the product rule [MATH]e^{2x} \cdot e^{\sin(4x)} \cdot 4\cos(4x) + e^{\sin(4x)} \cdot 2e^{2x} = 2e^{2x} \cdot e^{\sin(4x)}[2\cos(4x)+1][/MATH] Click to expand... Ah I see! However, I still don't get exactly the same answer - please refer to the attachment.
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Apr 13, 2021 #4 they are the same expression... note that [MATH]2e^{2x} \cdot e^{\sin(4x)} = 2e^{2x+\sin(4x)}[/MATH]
E eutas1 Junior Member Joined Apr 8, 2021 Messages 165 Apr 13, 2021 #5 skeeter said: they are the same expression... note that [MATH]2e^{2x} \cdot e^{\sin(4x)} = 2e^{2x+\sin(4x)}[/MATH] Click to expand... Oh yeah... I should have expanded them to check. Thank you!
skeeter said: they are the same expression... note that [MATH]2e^{2x} \cdot e^{\sin(4x)} = 2e^{2x+\sin(4x)}[/MATH] Click to expand... Oh yeah... I should have expanded them to check. Thank you!
