What is the derivative of e^logx+1 ?

The derivative of e^logx+1 is e^lgx

What is the first derivative of e^logx+1 ?

The first derivative of e^logx+1 is e^lgx

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derivative of

e^{l} o gx + 1

e^{l} gx

Solution steps

\frac{d}{d o} (e^{l} o gx + 1)

Treat

l, g, x

as constants

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d o} (e^{l} o gx) + \frac{d}{d o} (1)

\frac{d}{d o} (e^{l} o gx) = gx e^{l}

\frac{d}{d o} (1) = 0

= gx e^{l} + 0

Simplify

= e^{l} gx

y = (2 t - 3)^{2}

\frac{2 x}{( 1 - x ^{2} ) ^{2}}

10 + x

1 2^{0.5} - x^{1.5}

- 5 x + 10

What is the derivative of e^logx+1 ?
The derivative of e^logx+1 is e^lgx
What is the first derivative of e^logx+1 ?
The first derivative of e^logx+1 is e^lgx