Explain why the function y=e^(x+c) has a derivative that is itself for all values of c.?

Question

Explain why the function y=e^(x+c) has a derivative that is itself for all values of c.

Answer 1

Because it is a constant multiple of y=e^x, with the coefficient in this case being e^c.

Answer 2

This can be done using the chain rule. According to the chain rule, if you have a composite function, f(g(x)), and both of the separate functions are differentiable, then so is f(g(x)). Let y = f(g(x)), and let u = g(x), then:

dy/dx = dy/du * du/dx

If f(g(x)) = e^(x+c), let u = (x+c). Then the derivative of e^(x+c) is:

d/dx e^(x+c)

= d/dx e^(x+c) * d/dx (x+c)

= e^(x+c)

Answer 3

dy=d(e^(x+c))
=(e^(x+c))*d(x+c)
=(e^(x+c))*d(x) + (e^(x+c))*d(c)

d(c)=0 for any value of c
d(x)=1

so dy=(e^(x+c))*d(x) + (e^(x+c))*d(c)
=(e^(x+c))*(1) + (e^(x+c))*(0)
=e^(x+c)

therefore dy=y for all values of c