Derivative of $y=(1-\sin x)^e$?

Question

$$\begin{align} y&=(1-\sin x)^e\\ \ln y&=e\ln(1-\sin x)\\ \ln y&=e\ln1-e\ln(\sin x)\\ \ln y&=0-e\ln(\sin x)\\ 1/y&=e\ln(\sin x)'\\ 1/y&=e'\ln(\sin x)+e\ln(\sin x)'\\ 1/y&=0+\cos x/\sin x\\ y&=\tan x \end{align}$$ I am 100% sure I have made mistakes.

Answer 1

Most of the other answers deal with a computational error that you have made. I would like to deal with a notational (or perhaps communication) error that exists in your solution. You write

\begin{align} y&=(1-\sin x)^e\\ \ln y&=e\ln(1-\sin x)\\ \ln y&=e\ln1-e\ln(\sin x)\\ \ln y&=0-e\ln(\sin x)\\ 1/y&=e\ln(\sin x)'\\ 1/y&=e'\ln(\sin x)+e\ln(\sin x)'\\ 1/y&=0+\cos x/\sin x\\ y&=\tan x \end{align}

While I understand what you are attempting to communicate, the way that you have written it is "ungrammatical" at best, and bordering on nonsensical. Note that this isn't your fault—we (as in "the mathematical community at large") don't generally do a very good job of teaching students how to write mathematics before graduate school.

In mathematical writing (as in all writing), ideas should be structured in such a way to make clear what you are trying to say. Here, this would involve some notation which indicates how the lines of your argument should be read. Does each line imply the succeeding line? Are these if-and-only-if statements? or is this a system of equations that we want to see satisfied?

Were I to write out the argument (making the corrections that have been noted in several of the other answers), I might write

Our goal is to find $y'$, where $y := (1-\sin(x))^{\mathrm{e}}$. This can be done via implicit differentiation as follows: \begin{align} y = (1-\sin(x))^{\mathrm{e}} &\implies \ln(y) = \mathrm{e} \ln(1-\sin(x)) \\ &\implies \frac{y'}{y} = \mathrm{e} \frac{1}{1-\sin(x)} \frac{\mathrm{d}}{\mathrm{d}x}(1-\sin(x)) = -\frac{\mathrm{e} \cos(x)}{1-\sin(x)} \tag{$ \ast$} \\ &\implies y' = y \left( -\frac{\mathrm{e}\cos(x)}{1-\sin(x)} \right). \end{align} Then, since $y$ is explicitly given as a function of $x$ above, we can substitute in order to obtain $$ y' = (\color{blue}{1-\sin(x)})^{\mathrm{e}}\left( -\frac{\mathrm{e}\cos(x)}{\color{blue}{1-\sin(x)}} \right) = -\mathrm{e}(1-\sin(x))^{\mathrm{e}-1}\cos(x). $$

At ($\ast$) note that (as others have pointed out), we have $$ \frac{\mathrm{d}}{\mathrm{d}x} \ln(y) = \frac{1}{y}\cdot \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y'}{y}. $$ This follows from the chain rule, as $y$ is a function of $x$. Also, I'll admit that I sometimes have a tendency to be a bit verbose—you could almost certainly remove all of the talky bits (i.e. the sentences at the beginning and in the middle of the displayed math) and join the two bits of displayed math with an implication, and still be perfectly well understood, i.e. you could simply write

\begin{align} y = (1-\sin(x))^{\mathrm{e}} &\implies \ln(y) = \mathrm{e} \ln(1-\sin(x)) \\ &\implies \frac{y'}{y} = \mathrm{e} \frac{1}{1-\sin(x)} \frac{\mathrm{d}}{\mathrm{d}x}(1-\sin(x)) = -\frac{\mathrm{e} \cos(x)}{1-\sin(x)} \\ &\implies y' = y \left( -\frac{\mathrm{e}\cos(x)}{1-\sin(x)} \right) = -\mathrm{e}(1-\sin(x))^{\mathrm{e}-1}\cos(x). \end{align}

Of course, if you already happen to know the power rule in it's full generality; i.e. you know that $$ \frac{\mathrm{d}}{\mathrm{d}x} x^a = a x^{a-1} $$ for any $a \in \mathbb{R}$ (as long as $x > 0$), then the entire computation is much easier (again using chain rule):

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x} (1-\sin(x))^{\mathrm{e}} = \mathrm{e} (1-\sin(x))^{\mathrm{e}-1} \frac{\mathrm{d}}{\mathrm{d}x} (1-\sin(x)) = \mathrm{e} (1-\sin(x))^{\mathrm{e}-1} (-\cos(x)), $$

which gives the same answer in a single line of computation. $\ddot \smile$

Answer 2

$$y'=e(1-\sin x)'(1-\sin x)^{e-1}=-e\cos x(1-\sin x)^{e-1}.$$

Answer 3

The main mistake is the the logarithm is not additive, and that you differentiated $\ln y$ wrong. And you don't really need it here, you usually use logarithmic differentiation when the exponent depends on $x$ (would you use it to differentiate $y=x^2$?).

Here, using the Chain Rule, $$ y'=e(1-\sin x)^{e-1}\,(1-\sin x)'=-e\cos x\,(1-\sin x)^{e-1}. $$ And, although not related, $\tan x=\frac{\sin x}{\cos x}$.

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If you want to do it by logarithmic differentiation, you have $\ln y=e\,\ln(1-\sin x)$, so $$ \frac{y'}y=-e\,\frac{\cos x}{1-\sin x}, $$ so $$ y'=-e\,\frac{y \cos x}{1-\sin x}=-e\,\cos x\,(1-\sin x)^{e-1}. $$

Answer 4

HINT

Let use

$$([f(x)]^a)’=a[f(x)]^{a-1}f’(x)$$

Answer 5

we get $$\ln(y)=e\ln(1-\sin(x))$$ so $$\frac{y'}{y}=\frac{-e\cos(x)}{1-\sin(x)}$$

Answer 6

Your mistake was writing $\ln(1-\sin(x)) = \ln(1)-\ln(\sin(x))$ . thats wrong.

You have $u =(1-\sin(x))^e$

using the chain rule gives,$\quad$ $u' = e(1-\sin x)'(1-\sin x)^{e-1}=-e\cos x(1-\sin x)^{e-1}$

Answer 7

$$y=e(1-\sin(x))^{e-1}(1-\sin(x))'$$

Answer 8

Not a million mistakes...

$$\begin{align}y=(1-\sin x)^e\\ \ln y=e\ln(1-\sin x) \end{align} $$ You made a mistake here $\ln(a+b)\neq \ln a + \ln b$

Maybe you confused with $\ln(ab)=\ln a + \ln b$

$$\begin{align}\ln y=e\ln1-e\ln(\sin x)\\ \ln y=0-e\ln(\sin x)\\ 1/y=e\ln(\sin x)'\end{align} $$ Another mistake here $(\ln y)'= \frac {y'}y$ $$\begin{align} 1/y=e'\ln(\sin x)+e\ln(\sin x)'\\ 1/y=0+\cos x/\sin x\\ y=tgx \end{align} $$

Answer 9

There are two ways of doing this.

First,

$e$ is a constant. This means that the exponent rule applies here. Combining that with the chain rule yields

$$y = (1-\sin(x))^e \rightarrow y’ = e \cdot (1-\sin(x))^{e-1} \cdot (-\cos(x)).$$

Second,

we may take the logarithm of both sides to find another convenient form and then use the various derivative rules:

$$y = (1-\sin(x))^e$$

$$\Rightarrow \ln(y) = e \cdot \ln(1-\sin(x))$$

$$\Rightarrow \frac{y’}{y} = e \cdot \frac{1}{1-\sin(x)} \cdot (-\cos(x))$$

$$\Rightarrow y’ = e \cdot \frac{y}{1- \sin(x)} \cdot (-\cos(x))$$

$$\Rightarrow y’ = e \cdot (1-\sin(x))^{e-1} \cdot (-\cos(x)).$$