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

4 Answers
Apr 4, 2018

dy/dx = 5^x log5

Explanation:

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

Apr 4, 2018

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]

Apr 4, 2018

f'(x)=ln5*5^x

Explanation:

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

Apr 4, 2018

5^x*ln5

Explanation:

Given: f(x)=5^x

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

f'(x)=5^x*ln5