Find dy/dx by implicit differentiation and simplify the result
1. x=sec(1/y)
2. 4xy+ln(x^2y)=7
Hello I sort of know how to do these problems but not really. :( Can someone show all the work to solving them it would help me greatly. Thank you a lot.
x=sec(1/y)
1 = sec(1/y)tan(1/y)(-1/y^2) dy/dx
so,
dy/dx = -y^2 cos(1/y) cot(1/y)
or,
x=sec(1/y)
1/y = arcsec(x)
y = 1/arcsec(x)
y' = -1/[x√(x^2-1) arcsec(x)^2]
you can see that they are the same, since
cos(arcsec(u)) = 1/u
tan(arcsec(u)) = √(u^2-1)
4xy+ln(x^2y)=7
4y + 4xy' + 1/(x^2y)(2xy+x^2y') = 0
4xy' + 1/y y' = -4y - 2/x
y' = -2(2y + 1/x)/(4x + 1/y)
=
-2y(2xy+1)
---------------
x(4xy+1)
or, note that
4xy+ln(x^2y)=7
4xy + 2lnx + lny = 7
4y + 4xy' + 2/x + 1/y y' = 0
(4x+1/y)y' = -4y - 2/x
which works out the same
let z = 1/y
dx = d/dz sec(z)dz
dx = sec (z) tan (z) dz
but dz = (-1/y^2)dy
so
dx = sec(1/y) tan(1/y) (-1/y^2) dy
now
dy/dx = -y^2/[ sec(1/y) tan(1/y]
Sure! I can help you with these problems. I'll go through each problem step by step and explain how to use implicit differentiation to find dy/dx.
1. x = sec(1/y):
To differentiate implicitly, we treat y as a function of x, i.e., y = y(x). Then we differentiate both sides with respect to x, using the chain rule where necessary.
Differentiating x = sec(1/y) with respect to x:
d/dx(x) = d/dx(sec(1/y))
On the left side, d/dx(x) is simply 1. On the right side, we need to apply the chain rule. Recall that the derivative of sec(u) is sec(u)tan(u), and the derivative of 1/y with respect to x is -1/y^2 dy/dx.
So, we have:
1 = sec(1/y)tan(1/y)(-1/y^2) dy/dx
Simplifying this equation:
dy/dx = -y^2sec(1/y)tan(1/y)
Therefore, dy/dx for the given equation is -y^2sec(1/y)tan(1/y).
2. 4xy + ln(x^2y) = 7:
Again, we'll differentiate both sides of the equation with respect to x using the chain rule where necessary.
Differentiating 4xy + ln(x^2y) = 7 with respect to x:
d/dx(4xy) + d/dx(ln(x^2y)) = d/dx(7)
Now let's differentiate each term separately:
For the term 4xy, we use the product rule. The derivative of 4xy with respect to x is:
4y + 4x(dy/dx)
For the term ln(x^2y), we use the chain rule. The derivative of ln(u) is 1/u times the derivative of u. Therefore, the derivative of ln(x^2y) with respect to x is:
1/(x^2y) * d/dx(x^2y)
Simplifying this expression:
(1/(x^2y)) * d/dx(x^2y) = (1/(x^2y)) * (2xy + x^2(dy/dx))
On the right side of the equation, d/dx(7) is simply zero.
Putting it all together, our equation becomes:
4y + 4x(dy/dx) + (1/(x^2y)) * (2xy + x^2(dy/dx)) = 0
Now we can solve for dy/dx:
4y + (1/(x^2y)) * (2xy + x^2(dy/dx)) = -4x(dy/dx)
Bringing the terms involving dy/dx to one side:
4y - 4x(dy/dx) - (2xy/(x^2y)) = -x^2(dy/dx)
Simplifying further:
4y - 4x(dy/dx) - 2/x = -x^2(dy/dx)
Now we can isolate the dy/dx term:
-4x(dy/dx) + x^2(dy/dx) = -4y - 2/x
Combining like terms:
(-4x + x^2)(dy/dx) = -4y - 2/x
Finally, we can solve for dy/dx:
dy/dx = (-4y - 2/x) / (-4x + x^2)
Therefore, dy/dx for the given equation is (-4y - 2/x) / (-4x + x^2).
I hope this helps! Let me know if you have any other questions.