# Y=(x^5)(5^x)

 Anonymous From: -Posts: -Votes: - Y=(x^5)(5^x) Anonymous From: -Posts: -Votes: - D/dx(y(x)) = d/dx(5^x x^5) | The derivative of y(x) is y'(x): = | y'(x) = d/dx(5^x x^5) | Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = 5^x and v = x^5: = | y'(x) = x^5 (d/dx(5^x))+5^x (d/dx(x^5)) | The derivative of 5^x is 5^x log(5): = | y'(x) = 5^x (d/dx(x^5))+x^5 (5^x log(5)) | The derivative of x^5 is 5 x^4: = | y'(x) = 5^x x^5 log(5)+5^x (5 x^4)  Anonymous From: -Posts: -Votes: - Lny = 5lnx + xln5 y'/y = 5/x + ln5 y' = (x5)(5x)[5/x + ln5]  Anonymous From: -Posts: -Votes: - It's very simple. All you have to do is select any number you want and substitute on the "x". That's your "x" value. Solve for "y" and you have your "y" value. When you have the "x" and "y" values you have a point in a graph called (x, y). For example, lets make our x = 0. So, y = 5x y = 5(0) y = 0 So, when x = 0, y = 0. You have your point: (0, 0). Try another number. How about x = 1? y = 5x y = 5(1) y = 5. So, when x = 1, y = 5. That's your second point: (1, 5) Now, I would suggest doing at least two more of these (so that you have at least four points in the graph - this makes for a much more accurate graph than if you had just the two points (but you CAN graph only two points). Mark them down on a graph paper and then connect the dots!  Anonymous From: -Posts: -Votes: - 5X5*5X5 = 625X^2  Anonymous From: -Posts: -Votes: - Solution: lets: x = 0. So, y = 5x y = 5(0) y = 0 So, when x = 0, y = 0. You have (0, 0). if, x = 1? y = 5x y = 5(1) y = 5. So, when x = 1, y = 5. we have (1, 5)  Enter 