How do you differentiate #y=b^x?

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Answer 1 · 2016-07-30T08:27:43

#dy/dx=b^x *ln b#

From the given #y=b^x#

#ln y=ln b^x#

#ln y=x*ln b#

#d/dx(ln y)=d/dx(x*ln b)#

#(1/y)*y'=(x*0+ln b)#

#y'=y*ln b#

#y'=b^x*ln b#

God bless.....I hope the explanation is useful.