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Differentiate

$$ (1) \log y=e^{x}+4$$ $$(2) \frac{1}{2^{y}}=\frac{1}{2^{x}}+5$$

Please write full steps and if possible give an explantion. Thank You.

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1  
How to ask a homework question? –  Gigili May 26 '12 at 4:56
    
@user32251: Please see meta.math.stackexchange.com/questions/1803/… –  user9413 May 26 '12 at 4:56
    
You already have a hint below. You should at least try to solve it yourself. –  Eugene May 26 '12 at 5:11
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It seems an oddly worded question. One doesn't really differentiate an equation, one differentiates a function. I would have expected something like "If $\log y=e^x+4$, find $\frac{dy}{dx}$." –  André Nicolas May 26 '12 at 5:20
    
Editing many times by different users has changed the question! I can't figure out what was wrong with the first version. –  Gigili May 26 '12 at 5:31

3 Answers 3

up vote 3 down vote accepted

$(1)$

There are two ways to solve the first one. If you are familiar with implicit differentiation then differentiate both sides to get $$\begin{align*} \frac{d}{dx}(\log y)&=\frac{d}{dx}(e^x+4)\\ \frac{1}{y}\frac{dy}{dx}&=e^x+0\\ \frac{dy}{dx}&=ye^x\end{align*}$$ If you are not familiar with implicit differentiation then write $\log y=e^x+4$ as $$y=e^{e^x+4}$$ Differentiate and get $$\begin{align*} &\frac{dy}{dx}=\frac{d}{dx}\left(e^{e^x+4}\right)\\ &\frac{dy}{dx}=e^x\cdot e^{e^x+4}=ye^x\end{align*}$$


(2)

In the second question we can also use implicit differentiation or explicitly solve for y. Using implicit we get $$\begin{align*} \frac{1}{2^y}&=\frac{1}{2^x}+5\\ \frac{d}{dx}\left(\frac{1}{2^y}\right)&=\frac{d}{dx}\left(\frac{1}{2^x}+5\right)\\ \frac{-\ln2}{2^y}\frac{dy}{dx}&=\frac{-\ln2}{2^x}\\ \frac{dy}{dx}&=\frac{2^y}{2^x}\\ \end{align*}$$ But since $2^y=\frac{1}{\frac{1}{2^x}+5}$ $$\frac{dy}{dx}=\frac{\frac{1}{\frac{1}{2^x}+5}}{2^x}=\frac{1}{1+5\cdot2^x}$$ The other approach is by explicitly solving for $y$ and then differentiating as follows: $$\begin{align*} \frac{1}{2^y}&=\frac{1}{2^x}+5\\ \Longrightarrow2^y&=\frac{1}{\frac{1}{2^x}+5}\\ \Longrightarrow\log_2(2^y)&=\log_2\left(\frac{1}{\frac{1}{2^x}+5}\right)\\ \Longrightarrow y&=\log_2 1-\log_2(\frac{1}{2^x}+5)\\ \Longrightarrow y&=-\frac{\ln(\frac{1}{2^x}+5)}{\ln2}\\ \frac{dy}{dx}&=\frac{-1}{\ln2}\frac{d}{dx}\left(\ln\left(\frac{1}{2^x}+5\right)\right)\\ \frac{dy}{dx}&=\frac{-1}{\ln2}\cdot\frac{1}{\frac{1}{2^x}+5}\cdot \frac{-\ln2}{2^x}\\ \frac{dy}{dx}&=\frac{1}{1+5\cdot2^x} \end{align*}$$

*Remember that $\displaystyle\log_b a= \frac{\log_c a}{\log_c b}$

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You Saw wrong second question. –  Neer May 26 '12 at 5:40
    
@Neer I fixed it now –  E.O. May 26 '12 at 6:10
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@Neer This is a rather poor comment to someone who wrote you a very long and detailed answer to your badly-written question. –  Phira May 26 '12 at 11:43
    
Sorry and thank you @E.O. –  Neer Apr 6 '13 at 9:17

$$\log y= e^{x} +4$$ $$ e^{\log y}= e^{(e^{x}+4)}$$ $$\Rightarrow y= e^{(e^{x}+4)}$$ Now differntiate with respect to $x$, we get, $$ \frac{dy}{dx}= e^{(e^{x}+4)}.e^{x}$$

$$\frac{1}{2^{y}}=\frac{1}{2^{x}+5}$$ $$ \therefore 2^{y}= 2^{x}+5$$ $$ \log{2^{y}}=\log{(2^{x}+5)}$$ $$\Rightarrow y\log 2= \log{(2^{x}+5)}$$ Now differentiate with respect to $x$, we get, $$\log 2\frac{dy}{dx}= \frac{1}{2^{x}+5}.2^{x}\log 2$$ $$\therefore \frac{dy}{dx}=\frac{2^{x}}{2^{x}+5}$$

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Second question has changed. –  Neer May 26 '12 at 5:23

For the first one, you need to exponentiation both sides of the equation to get $y = e^{(4+e^x)}$ (assuming you are finding $dy/dx$), then go to the following for an exact step-by-step solution:
Wolfram|Alpha1

For the second one, the first step is to re-arrange it as: $$\frac{1}{ 5+\frac{1}{2^x}} = 2^y$$ Then take the base to logarithm of each side to get:

$$y = \log_2 \frac{1}{ 5+\frac{1}{2^x}}$$

then to compute the derivative go to the following:

Wolfram|Alpha2

Wolfram Alpha is a great tool for little pesky problems like this.... but don't overdo it!

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If you use the hyperlink tool above the edit box, it usually takes care of the escaping correctly so that your link does what it should. –  Dylan Moreland May 26 '12 at 5:07

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