How do I find #f'(x)# for #f(x)=5^x# ?

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Answer 1 · 2018-04-04T06:18:09

#dy/dx = 5^x log5 #

#y=5^x#

#logy=x log5#

#dy/dx. 1/y=x.(0)+log5.(1)#

#dy/dx . 1/y =log5#

#dy/dx =log5 . y#

#dy/dx =log5 . 5^x#

#dy/dx = 5^x log5 #

Answer 2 · 2018-04-04T06:21:23

#f(x)=y=5^x#

Taking log on both the sides,

#logy=log5^x#

#logy=xlog5# #color(white)(wwggggggggggggw# #["as " color(red)(loga^b = bloga)]#

Now, applying implicit differentiation,
#1/y dy/dx=x (d(log5))/dx + log5# #color(white)(o# #["product rule is applied on "xlog5 ]#

# dy/dx=y[x + log5]# #color(white)(gggggggw# #["as derivative of a constant is 1" ]#

# dy/dx=5^x[x + log5]#

Answer 3 · 2018-04-04T06:27:56

#f'(x)=ln5*5^x#

Let #y=f(x)=5^x#

then #lny=xln5# and differentiating we get

#1/y(dy)/(dx)=ln5#

or #(dy)/(dx)=ln5*y=ln5*5^x#

Answer 4 · 2018-04-04T14:00:20

#5^x*ln5#

Given: #f(x)=5^x#

Using the common integral that #(a^x)'=a^xlna#, we get:

#f'(x)=5^x*ln5#