Differentiate $y = 2^x + 2^{-x}$

Question

I have a question about negative powers in implicit differentiation.

I was asked to differentiate the equation

$$y = 2^x + 2^{-x}$$

Now, from previous studies, I can prove the following:

$y = a^x$

$\ln y = \ln a^x = x \ln a$

$\frac{1}{y} \frac{dy}{dx} = \ln a$

$\frac{dy}{dx} = y \ln a = a^x \ln a$

hence differentiating $y = a^x$ results as $\frac{dy}{dx} = a^x \ln a$.

Using the above information, I attempted to differentiate the equation above via implicit differentiation:

$$\frac{dy}{dx} = 2^x \ln 2 + 2^{-x} \ln 2$$

However, apparently, my solution is incorrect. Instead, it is

$$\frac{dy}{dx} = 2^x \ln 2 - 2^{-x} \ln 2$$

I'm confused. Why is my solution wrong?

Answer 1

$y = a^{-x}$

$\ln y = \ln a^{-x} = -x \ln a$

$\frac{1}{y} \frac{dy}{dx} = -\ln a$

$\frac{dy}{dx} =- y \ln a = -a^{-x} \ln a$

hence when differentiating $y = a^{-x}$ results as $\frac{dy}{dx} = -a^{-x} \ln a$.

credit - @MANMAID, @smurf

Answer 2

You can differentiate $2^{-x}$ in two ways (at least).

You can use the chain rule. This rule, roughly speaking, says that the derivative of $f(g(x))$ is $g'(x)f'(g(x))$. To apply this rule to $2^{-x}$ let $f(x)=2^x$ and $g(x)=-x$. Then $f(g(x))=2^{-x}$ and the derivative is $g'(x)f'(g(x))=-1\cdot\ln 2\cdot 2^{g(x)}=-\ln2\cdot 2^{-x}$.

The other way is as follows. Write $2^{-x}=\left(\frac12\right)^x$. Then, its derivative is $\ln(1/2)\left(\frac12\right)^x=-\ln 2\cdot 2^{-x}$.

Actually, there are other ways, including implicit differentiation and differentiaitng $1/2^x$ as a quotient of functions. But I think that they are more complicated.

Answer 3

Let $u=-x,$ so that $y = 2^{-x} = 2^u.$

The chain rule tells you that $$ \frac d {dx} 2^{-x} = \frac{dy}{dx} = \frac {dy}{du}\cdot\frac{du}{dx} = \left( \frac d {du} 2^u\right) \cdot \frac d {dx} (-x) = \cdots\cdots. $$