Find the derivative of$\ (2^x+1)^{\frac{1}{2}} \ $

Question

I have a question for my math course but I'm not sure if I should use the chain rule to solve it. The question is find the derivative of $(2^x +1)^{1/2}.$ I did $(1/2)(2^x+1)^{1/2-1} 2\ln 2$ to get $(\ln 2)(2^x+1)^{-1/2}$ but I'm not sure if it's right. Any help would be appreciated!

Answer 1

Using the power rule and the chain rule we get $$\frac{1}{2}(2^x+1)^{-1/2}\cdot 2^x\ln(2)$$

Answer 2

The derivative of $\ (2^x+1)^{\frac{1}{2}} \ $is as follows:

$$\frac{d}{dx}(2^x+1)^{\frac{1}{2}} \ $$ $$\frac{1}{2}(2^x+1)^{\frac{1}{2}-1}\cdot (\ln(2))2^{x} \ $$ $$\frac{1}{2}(2^x+1)^{\frac{1}{2}-\frac{2}{2}}\cdot (\ln(2))2^{x} \ $$ $$\frac{1}{2}(2^x+1)^{-\frac{1}{2}}\cdot (\ln(2))2^{x} \ $$ $$\frac{1}{2\sqrt{2^{x}+1}}\cdot (\ln(2))2^{x} \ $$ $$\frac{(\ln(2))2^{x}}{2^{1}\sqrt{2^{x}+1}} \ $$ $$\frac{(\ln(2))2^{x-1}}{\sqrt{2^{x}+1}} \ $$

I hope this helps!

Answer 3

$$(\sqrt{t+1})'=\frac{t'}{2\sqrt{t+1}}$$ and

$$(\sqrt{2^x+1})'=\frac{2^x\ln 2}{2\sqrt{2^x+1}}$$

Answer 4

Hint: Problem comes with $2^x$, so, write $e^{ln(2)x}$ instead