# Find the equation of the tangent to the curve $y = {2^x} + {2^{ - x}}$ at the point $(2,4{1 \over 4})$

\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!

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

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: $$\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)$$

• $ln(y) = ln({2^x} + {2^{ - x}})$, could you explain why this is? I dont want to make the same mistake again.. Thank you – seeker Apr 30 '13 at 1:45
• @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 $\ln(a+b)=\ln(a)+\ln(b)$. – Scott H. Apr 30 '13 at 1:48

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

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

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

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

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?

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 $n=\frac{17}{4} - (\ln2\cdot\frac{17}{4})\cdot 2$