X^4 ?
4x^3
f'(x) = x^(3) / 3 Hence f(x) = 3x^(2) / 3 +C f(x) = x^(2) + C ' The '3' cancels out. The 'C' is a constant that comes with antoderivation.
A dot A = A2 do a derivative of both sides derivative (A) dot A + A dot derivative(A) =0 2(derivative (A) dot A)=0 (derivative (A) dot A)=0 A * derivative (A) * cos (theta) =0 => theta =90 A and derivative (A) are perpendicular
The derivative of e7x is e7 or 7e.The derivative of e7x is 7e7xThe derivative of e7x is e7xln(7)
the derivative is 0. the derivative of a constant is always 0.
Derivative of 4x is 4.
The derivative of x^n is nx^(n-1) any n. The derivative of x^4 is 4x^3.
0.3333
d/dx(x4/4) = x3
ln(x4)?d/dx(ln(u))=1/u*d/dx(u)d/dx(ln(x4))=[1/x4]*d/dx(x4)-The derivative of x4 is:d/dx(x4)=4x4-1d/dx(x4)=4x3d/dx(ln(x4))=[1/x4]*(4x3)d/dx(ln(x4))=4x3/x4d/dx(ln(x4))=4/x(lnx)4?Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(lnx)4=4(lnx)4-1*d/dx(lnx)d/dx(lnx)4=4(lnx)3*d/dx(lnx)-The derivative of lnx is:d/dx(ln(u))=1/u*d/dx(u)d/dx(lnx)=1/x*d/dx(x)d/dx(lnx)=1/x*(1)d/dx(lnx)=1/xd/dx(lnx)4=4(lnx)3*(1/x)d/dx(lnx)4=4(lnx)3/x
10/x3 = 10 x-3d/dx(10x-3) = -30 x-4 = -30/x4
If you mean: f(x) = x4 - 3x3 + 5x2 / x2 then: f(x) = x4 - 3x3 + 5 ∴ f'(x) = 4x3 - 9x2 If you mean: f(x) = (x4 - 3x3 + 5x2) / x2 then: f(x) = x2 - 3x + 5 ∴ f'(x) = 2x - 3
X4.
f'(x) = x^(3) / 3 Hence f(x) = 3x^(2) / 3 +C f(x) = x^(2) + C ' The '3' cancels out. The 'C' is a constant that comes with antoderivation.
"Derivative of"
1
(x4 - 3)(x4 + 3)
well, the second derivative is the derivative of the first derivative. so, the 2nd derivative of a function's indefinite integral is the derivative of the derivative of the function's indefinite integral. the derivative of a function's indefinite integral is the function, so the 2nd derivative of a function's indefinite integral is the derivative of the function.