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

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

Explain why the function y=e^(x+c) has a derivative that is itself for all values of c.?
Because it is a constant multiple of y=e^x, with the coefficient in this case being e^c.
Reply: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)
Reply: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