What is the derivative of #y# with respect to #x# for #y = 17^x#?

Question

10949 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-26T12:15:13

The derivative is #f'(x)=dy/dx=d17^x/dx=17^x*ln17#

The general rule of differentiating exponential functions is that:

#(a^x)'=(a^x)*lna#

Answer 2 · 2015-12-30T07:13:09

#y'=17^x*ln17#

The first step is to rewrite #17^x# as something differentiable.

#y=17^x=e^(ln17^x)=e^(xln17)#

#e^(xln17) # is differentiable since we can use the chain rule:

#d/dx(e^u)=e^u*u'#

Apply this to #e^(xln17)#:

#y'=d/dx(e^(xln17))=e^(xln17)*d/dx(xln17)#

Two things to consider here:

#color(white)(sss)# #e^(xln17)# is still equal to #17^x#, we can rewrite it as such now that
#color(white)(sss)# we've differentiated.

#color(white)(sss)# #d/dx(xln17)=ln17#
#color(white)(sss)# Remember, #ln17# is just a constant. Don't be fooled by it.

Thus,

#y'=17^x*ln17#