Tricky differentiation

Calculus Level 2

$\sin ( x + y ) = \log ( x + y )$

Find the value of $\frac {dy}{dx}$ .

Note that we didn't give you the base of the logarithm, you should not specifically assume that the logarithm is in base $e$ or 10.

The answer is -1.

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Calvin Lin Staff
Feb 2, 2015

Differentiate implicitly with respect to y, we have

$(\cos ( x + y) - \frac{1}{x+y} ) ( 1 + \frac{dy}{dx} ) = 0.$

To solve $\cos ( x + y ) - \frac{1}{x+y} = 0$ , let $x + y = t$ . Then, we have $\cos t = \frac{1}{t}$ . We claim that the solution set consists of isolated points, which you can show by differentiating. Thus $x+y =t$ is a constant and so $\frac{dy}{dx} = - 1$ .

To solve the second factor, we get $\frac{dy}{dx} = -1$ immediately.

Shouldn't the question be $\ln(x+y)$

I thought the derivative of $\log(x)$ is $\frac{1}{x\ln(10)}$

Figel Ilham - 6 years, 4 months ago

That is a valid point, but it doesn't affect this solution. The solution set of $\cos t = \frac{ k} { t}$ is going to be a discrete set for any value of $k$ . This means that $x + y$ in the neighborhood of a point must be a constant.

Because it's a calculus question, I almost always assume that they mean $\ln$ , unless there is evidence to the contrary.

Calvin Lin Staff - 6 years, 4 months ago

why not just solve for dy/dx algebraically in that first equation. you get dy/dx = -(cos(x+y)+ 1/(x+y)) / (cos(x+y)+1/(x+y) = -1

no need to do anything fancy with t as a parameter.

Aren Nercessian - 6 years, 4 months ago

Yes, that is what I did too, if you look at the first line. However, the concern here, is that you need to deal with the case where $\cos ( x + t) - \frac{1}{ x+ y } = 0$ (slight error in your working), since otherwise $\frac{0}{0} \neq 1$ .

You simply assumed that was not possible, which may not be the case. The thing is that, even in those cases, we will still have $\frac{ dy}{dx} = - 1$ .

Calvin Lin Staff - 6 years, 4 months ago

Jakub Šafin
Feb 1, 2015

Log is increasing from $-\infty$ to $\infty$ . Sin oscillates between -1 and 1. Therefore, there will be some value of $x+y=c$ such that $\sin{c}=\log{c}$ . Then, it's clear that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(c-x)=-1$ .

The important aspect is that the set of solution to $\sin c = \log c$ is a discrete set, which allows you to conclude that in a neighborhood of the solution point, $c=x + y$ is a constant.

Calvin Lin Staff - 6 years, 4 months ago

Tricky differentiation

The answer is -1.

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