Derivatives of Exponential Functions with Trignometric function power

What are exponential functions? A mathematical function which is in the form, f(x) = a^x is called an exponential function, here x is a variable and a is a constant which is the base of the function (a is greater than zero but does not equal one). The most commonly used exponential function is e which is the natural exponential function which is denoted as e^x. Some definitions of e are
• e=lim(n?inifinity)[1+1/n]^n
• Lim(h?0)[e^h-1]/h=1, where e is a unique positive number
• e=summation(n=0 to infinity)[1/n(factorial)]
We know that derivative is defined as, f’(x) = lim(h?0)[f(x+h) – f(x)]/h
Using the above, we can find the derivative of the natural exponential function f(x) = e^x
d[e^x]/dx = lim(h?0)[e^(h+x) – e^x]/h
    =lim(h?0)[e^h.e^x – e^x]/h  [using rule of exponents]
    =lim(h?0)e^x[e^h-1]/h            [taking e^x common]
    =e^x lim(h?0)[e^h-1]/h
    =e^x .1
    = e^x
As we observe, e^x is its own derivative. Let us find the derivative of the exponential function with trigonometric function as the power.

E-sinx
We have, derivative of e^u is given by e^u.du/dx. So, the derivative of E-sinx written as e^-sinx would be d[e^-sinx]/dx. Taking –sinx = u which gives du/dx = -cosx, we get, e^u. du/dx which equals e^-sinx.(-cosx)
Finally we get the derivative of E-sinx as –cosx.e^-sinx

E –sinx
We have, y= e^-sinx
     y= e^u where u= -sinx, derivative of u= du/dx = -cosx
taking derivative, we get,
dy/dx= dy/du. du/dx
          = e^-sinx. [-cosx]
          = -cosx.e^-sinx

E^-sinx
Let us find the derivative of the exponential function with the trigonometric function –sinx as the power.
Using the chain rule, we get,
d[e^(-sinx)]/dx = d[e^(-sinx)/dx. d[-sinx]/dx
             = e^(-sinx). (-cosx)
             = -cosx.e^(-sinx)