First, recall that
#d/dx(a^x)=a^xln(a)#. I'll show a proof for this at the end of the answer. Keeping this in mind, let's continue:
#f'(x)=d/dx10(5.1)^x+d/dx21(0.22)^x+d/dx5e^x#
(Differentiating a sum of various terms means we can break up the terms into individual derivatives.) (#d/dx(x_1+x_2)=d/dxx_1+d/dxx_2)#
#f'(x)=10d/dx(5.1)^x+21d/dx(0.22)^x+5d/dxe^x#
(We can always factor constants out of a derivative. In fact, it cleans up our work most of the time. #d/dxcx=cd/dxx#)
#f'(x)=10(ln(5.1))5.1^x+21(ln(0.22))0.22^x+5e^x#
The derivative of the natural exponential function #e^x# is just #e^x,# so nothing really changes there.
Proof of #d/dx(a^x)=a^xln(a)#:
#a^x=e^ln(a^x)# (From the fact that #x=e^ln(x)#)
#a^x=e^(xln(a))# (From the fact that #ln(a^x)=xln(a)#. Exponent rule for logarithms.)
#d/dxa^x=d/dxe^(xln(a))#
#d/dxe^(xln(a))=e^(xln(a))*d/dx(xln(a))=e^(xln(a))*ln(a)#
(From the fact that #d/dxe^f(x)=e^f(x)d/dxf(x)#. This is just the Chain Rule used to differentiate #e# raised to the power of some function #f(x).#)
#e^(xln(a))*ln(a)=a^xln(a)#
Recall that we originally got #e^(xln(a))# from rewriting #a^x# as #e^(ln(a^x))=e^(xln(a))#. We can just turn it back into an #a^x# at this point.)
Therefore, #d/dx(a^x)=a^xln(a)#
