1
$\begingroup$

$\eqalign{ & y = {2^x} + {2^{ - x}} \cr & \ln y = x\ln 2 - x\ln 2 \cr & \ln y = 0 \cr & {1 \over y}{{dy} \over {dx}} = 0 \cr & {{dy} \over {dx}} = 0 \cr} $

I've checked the answer and I've got the differential wrong, What am I doing wrong? I assume it has something to do with the expression with the negative exponent? Am I not allowed to prefix the natural expression with "minus" x? Could someone explain please, thank you!

$\endgroup$
  • 3
    $\begingroup$ $\log(2^x+2^{-x})\ne\log(2^x)+\log(2^{-x})$ $\endgroup$ – oldrinb Apr 30 '13 at 1:42
1
$\begingroup$

Your second step is incorrect: $$\ln(y)=\ln(2^x+2^{-x})$$ but $$\ln(y)\ne\ln(2^x)+\ln(2^{-x})$$ You don't need logs here, just differentiate each side as normal, using the following on the RHS: $$\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)$$

$\endgroup$
  • $\begingroup$ $ln(y) = ln({2^x} + {2^{ - x}})$, could you explain why this is? I dont want to make the same mistake again.. Thank you $\endgroup$ – seeker Apr 30 '13 at 1:45
  • 2
    $\begingroup$ @Assad If the LHS is equal to the RHS then their logs should be equal as well. We also know that $\ln(ab)=\ln(a)+\ln(b)$. However, its not the case that $\ln(a+b)=\ln(a)+\ln(b)$. $\endgroup$ – Scott H. Apr 30 '13 at 1:48
3
$\begingroup$

$$\ln(2^x+2^{-x}) \neq \ln(2^x)+\log(2^{-x})$$

Recall that $\ln(a+b) \neq \ln(a)+\ln(b)$. We do have that $\ln (ab) = \ln a + \ln b$

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

$$ y'=\dfrac{d}{dx}(2^x) + \dfrac{d}{dx}(2^{-x})\tag{1}$$

$$y' = 2^x \ln 2 + (-1)\cdot 2^{-x}\ln 2 \tag{2}$$ $$y' = 2^x \ln 2 - 2^{-x}\ln 2 \tag{3}$$

$$ y'= \left(2^x-2^{-x}\right)\ln 2\tag{4}$$

$(2)$ In general, if you have $$y = a^x;\;\; \text{then}\;\; y' =\dfrac{dy}{dx} = a^x\ln a\tag{a is constant}$$

$(3)$ Similarly, if you have $$y = a^{f(x)},\;\; \text{then}\;\; y'= \dfrac{dy}{dx} = a^{f(x)}f'(x)(\ln a)\qquad\tag{a is constant}$$

$\endgroup$
  • $\begingroup$ The two rules for differentiating (see the bottom two expressions) will come in very handy! $\endgroup$ – Namaste Apr 30 '13 at 2:05
  • $\begingroup$ Thank you @amwhy, you've made this much more lucid, thanks again! $\endgroup$ – seeker Apr 30 '13 at 2:06
2
$\begingroup$

$$y=2^x+2^{-x}\implies y'=2^x\log2-2^{-x}\log 2=\left(2^x-2^{-x}\right)\log 2$$


Neat off-topic stuff: consider $y=b^x+b^{-x}$ for some positive real $b$. Notice that differentiating yields $y'=(b^x-b^{-x})\log b$. In the case where $b=e$, we have $y=e^x+e^{-x}$ as well as $y'=e^x-e^{-x}$. Taking the second derivative gives us $y''=e^x+e^{-x}=y$ so we've found a solution to $y''-y=0$. What about $y=e^{ix}+e^{-ix}$? Isn't this a solution to $y''+y=0$? What other solutions to $y''-y=0$ do you know?

$\endgroup$
1
$\begingroup$

Again, no need to take logs.

$y'=\ln2(2^x+2^{-x})$. For $x=2$ this gives $y'=\ln2\cdot\frac{17}{4}$. This is your slope. Use this to solve for the intercept $n=\frac{17}{4} - (\ln2\cdot\frac{17}{4})\cdot 2$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.