Let
y = 3x
y' = 3(x)'
y' = 3(x1)'
y' = 3[1x(1-1)]
y' = 3(1x0)
y' = 3(1 x 1)
y' = 3
In general:
y = xn
y' nx(n-1)
The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.
If y = 3x +- 1, the derivative with respect to x is y' = 3.
3e3x
3sec2(3x)
the derivative of 3x is 3 the derivative of x cubed is 3 times x squared
The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.The derivate of 3x is 3; the derivative of -1 is 0. So, the derivative of 3x-1 is simply 3.
If y = 3x +- 1, the derivative with respect to x is y' = 3.
The derivative of y = sin(3x + 5) is 3cos(3x + 5) but only if x is measured in radians.
3e3x
3sec2(3x)
3x - 4 sqrt(2)The first derivative with respect to 'x' is 3.
the derivative of 3x is 3 the derivative of x cubed is 3 times x squared
9x2
The derivative of a cube function, such as ( f(x) = x^3 ), can be found using the power rule of differentiation. According to the power rule, the derivative ( f'(x) ) is given by ( 3x^{2} ). This means that the slope of the tangent line to the curve at any point ( x ) is ( 3x^{2} ). Therefore, the derivative of a cube, in this case, is ( 3x^{2} ).
In this case, you'll need to apply the chain rule, first taking the derivative of the tan function, and multiplying by the derivative of 3x: y = tan(3x) ∴ dy/dx = 3sec2(3x)
-1
d/dx 2x2+3x+7=4x+3