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_______________
dy
Find —— for an implicit function:
dx
cos(xy) = 3x + 1.
First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:
![\mathsf{\dfrac{d}{dx}\big[cos(xy)\big]=\dfrac{d}{dx}(3x+1)}\\\\\\ \mathsf{-\,sin(xy)\cdot \dfrac{d}{dx}(xy)=\dfrac{d}{dx}(3x)+\dfrac{d}{dx}(1)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bd%7D%7Bdx%7D%5Cbig%5Bcos%28xy%29%5Cbig%5D%3D%5Cdfrac%7Bd%7D%7Bdx%7D%283x%2B1%29%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B-%5C%2Csin%28xy%29%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdx%7D%28xy%29%3D%5Cdfrac%7Bd%7D%7Bdx%7D%283x%29%2B%5Cdfrac%7Bd%7D%7Bdx%7D%281%29%7D)
Apply the product rule to differentiate that term at the left-hand side:
![\mathsf{-\,sin(xy)\cdot \left[\dfrac{d}{dx}(x)\cdot y+x\cdot \dfrac{dy}{dx}\right]=3+0}\\\\\\ \mathsf{-\,sin(xy)\cdot \left[1\cdot y+x\cdot \dfrac{dy}{dx}\right]=3}\\\\\\ \mathsf{-\,sin(xy)\cdot \left[y+x\cdot \dfrac{dy}{dx}\right]=3}](https://tex.z-dn.net/?f=%5Cmathsf%7B-%5C%2Csin%28xy%29%5Ccdot%20%5Cleft%5B%5Cdfrac%7Bd%7D%7Bdx%7D%28x%29%5Ccdot%20y%2Bx%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%5D%3D3%2B0%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B-%5C%2Csin%28xy%29%5Ccdot%20%5Cleft%5B1%5Ccdot%20y%2Bx%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%5D%3D3%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B-%5C%2Csin%28xy%29%5Ccdot%20%5Cleft%5By%2Bx%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%5D%3D3%7D)
Now, multiply out the terms to get rid of the brackets at the left-hand
dy
side, and then isolate —— :
dx
and there it is.
I hope this helps. =)
Tags: <span><em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>
</span>
_______________
dy
Find —— for an implicit function:
dx
cos(xy) = 3x + 1.
First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:
Apply the product rule to differentiate that term at the left-hand side:
Now, multiply out the terms to get rid of the brackets at the left-hand
dy
side, and then isolate —— :
dx
and there it is.
I hope this helps. =)
Tags: <span><em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>
</span>
Answer: pages per hour.
Step-by-step explanation:
For this exercise it is important to know that, by definition, Direct variation equations have the following form:
Where "k" is the constant of variation.
Therefore, it is a straight line that passes through the origin.
In this case, given the graph attached in the exercise, you can follow these steps in order to find the constant of variation:
Step 1: You need to choose any point on the line. Let's choose the point (5,4).
Step 2: Now you must ubstitute the coordinates of that point into . Then:
Step 3: Finally, you must solve for the Constant of variation "k", in order to find its value. Therefore, you get that this is: