Question

Differentiate. f(x) = 6e-8x 0 48e-8x 0 6e-8x 0 -48e-8x 0 -8e-8x 1 -

Answer 1

Let f(x) = 6exp(-8x).

Using the properties of derivatives, differentiate f(x):

`(d)/(dx)f(x)=(d)/(dx)(6e^(-8x))`

Take the constant factor of 6 out of the derivative:

`(d)/(dx)(6e^(-8x))=6\cdot(d)/(dx)(e^(-8x))`

Let u = -8x and rewrite the expression:

`6\cdot(d)/(dx)(e^(-8x))=6\cdot(d)/(dx)(e^u)`

Using the Chain Rule, we know that:

`6\cdot(d)/(dx)(e^u)=6\cdot(d)/(du)(e^u)\cdot(d)/(dx)(u)`

The derivative of the exponential function e^u is again e^u:

`6\cdot(d)/(du)(e^u)\cdot(d)/(dx)(u)=6e^u\cdot(d)/(dx)(u)^{}`

Substitute back u = -8x :

`6\cdot(d)/(du)(e^u)\cdot(d)/(dx)(u)=6e^(-8x)\cdot(d)/(dx)(-8x)^{}`

Take the constant factor of -8 out of the derivative:

`6e^(-8x)\cdot(d)/(dx)(-8x)^{}=(-8)6e^(-8x)\cdot(d)/(dx)(x)^{}`

The derivative of x (with respect to x) is 1:

`(-8)6e^(-8x)\cdot(d)/(dx)(x)^{}=(-8)6e^(-8x)\cdot(1)`

Solve the corresponding products:

`(-8)6e^(-8x)\cdot(1)=-48e^(-8x)`

Therefore, the derivative of the function f(x) is given by:

`(d)/(dx)f(x)=-48e^(-8x)`