0
Still curious? Ask our experts.
Chat with our AI personalities
Lao
Beau
ViviAdd your answer:
Earn +20 pts
How do you integrate x power e power x?
{xe^x dx integrate by parts let f(x) = x so f'(x) = 1 and g(x) = e^x so g'(x) = e^x so.. f(x)*g(x) - {(g(x)*f'(x)) dx therefore xe^x - {(e^x * 1) dx so.. xe^x - e^x + C factorize so... (x-1)e^x + C
How do you integrate cot3x dx?
1/3ln(sin3x) + C
How you integrate xxxx 1 dx?
.2x^5+x+C
Why derivative of mod x does not exist graphically?
Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.
What is the derivative of f plus g of 3 and f times g of 3 given that f of 3 equals 5 d dx f of 3 equals 1.1 g of 3 equals -4 d dx g of 3 equals 7 Also please explain QUICK THANK YOU?
d/dx [f(x) + g(x)] = d/dx [f(x)] + d/dx [g(x)] or f'(x) + g'(x) when x = 3, d/dx [f(x) + g(x)] = f'(3) + g'(3) = 1.1 + 7 = 8.1 d/dx [f(x)*g(x)] = f(x)*d/dx[g(x)] + d/dx[f(x)]*g(x) when x = 3, d/dx [f(x)*g(x)] = f(3)*g'(3) + f'(3)*g(3) = 5*7 + 1.1*(-4) = 35 - 4.4 = 31.1
