I can see two likely intentions of this question. You wish to find the derivative of either:
y=7e-6x
or
y=7e-6x
For the first option, we use the fact that 7e is a constant (e is a number, not a variable, and a number times a number is still a number) to know that the derivative of the subcomponent 7e is zero. For the subcomponent -6x, we can use the power rule. Its derivative is (-6)(1)x1-1=-6x0=(-6)(1)=-6. So, in the end, we get:
y'=-6
For the second option, we use the chain rule. For a function that contains another function in itself, you must first derive the outer function and then multiply that by the derivative of the inner function. For this function, the "outer" function is ex, while the "inner" function is -6x. The derivative of ex is ex, and the derivative of -6x is -6. In the end, we get:
y'=(outer function derivative)(inner function derivative)=(7e-6x)(-6)=-42e-6x
y'=-42e-6x
Derivative of sin x = cos x, so chain rule to derive 8x = 8 , answer is 8cos8x
Because the derivative of e^x is e^x (the original function back again). This is the only function that has this behavior.
Calculate the derivative of the function.Use the derivative to calculate the slope at the specified point.Calculate the y-coordinate for the point.Use the formula for a line that has a specified slope and passes through a specified point.
If y = 3x +- 1, the derivative with respect to x is y' = 3.
y is 12
Find the derivative of Y and then divide that by the derivative of A
y=3 cos(x) y' = -3 sin(x)
y is a sum of constants and so is itself a constant. Its derivative is, therefore, zero.
Derivative of sin x = cos x, so chain rule to derive 8x = 8 , answer is 8cos8x
- the derivative with respect to x is 40y - The derivative with respect to Y is 40xSo, since both x and y equal 2, both derivatives yield 40*2 = 80
y"+y'=0 is a differential equation and mean the first derivative plus the second derivative =0.Look at e-x the first derivative is -e-xThe second derivative will be e-xThe sum will be 0
(xlnx)' = lnx + 1
y=(2x)2 y=2(2x) y=4x
if y=aex+be2x+ce3x then its derivative dy/dx or y' is: y'=aex+2be2x+3ce3x
Because the derivative of e^x is e^x (the original function back again). This is the only function that has this behavior.
Y = 36cot(x)Y' = dy/dx36cot(x)= - 36csc2(x)==========
D(y)= sin 2x