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.
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.
$(1)$
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 $\displaystyle\log_b a= \frac{\log_c a}{\log_c b}$
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 Alpha is a great tool for little pesky problems like this.... but don't overdo it!
$\log y= e^{x} +4$ $ 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, $\log 2\frac{dy}{dx}= \frac{1}{2^{x}+5}.2^{x}\log 2$ $\therefore \frac{dy}{dx}=\frac{2^{x}}{2^{x}+5}$