Question

Having a few problems needing help. Trying to Find dy/dx by implicit differentiation.

y4ex + x = y6x

Answer 1

take the derivative of each part as though it was a typical derivative, except that every time you take the derivative of y, you multiply that part by dy/dx, then once its all differentiated, put all the parts multiplied by dy/dx on one side of the equation and the non dy/dxs on the other side of the equal sign. then simplify so that you have dydx=______

Answer 2

ill do the one you posted so you can better understand

Answer 3

is y4 y^4?

Answer 4

??

Answer 5

yes, ok sorry about that

Answer 6

and is ex e^x?

Answer 7

I am not sure where i would find that?

Answer 8

here, tell me if this is the equation you are given:
\[y ^{4}e ^{x} +x = y ^{6}x\]
?

Answer 9

would dy=y 6 -y4ex-1

Answer 10

i see where your going, but that wont work in this case. an easier way would be to just go through it, is the equation i wrote above correct?

Answer 11

Yes the equation is correct...

Answer 12

Its just applying the equation with the problem always gets me

Answer 13

ok so then, and this applies to all equations, when you are using implicit differentiation, you can just do the following:

y4ex+x=y6x
(4y3ex(dy/dx)+exy4)+1=6y5x(dy/dx)+y6
i used the product rule twice there, and whenever i took the derivative of y, i multiplied it by "dy/dx" because we are differentiating in terms of x.

Answer 14

so then you take that
(4y3ex(dy/dx)+exy4)+1=6y5x(dy/dx)+y6
=
4y3ex(dy/dx)+exy4+1=6y5x(dy/dx)+y6

move the non dy/dx over

4y3ex(dy/dx)=6y5x(dy/dx)+y6-1-exy4

move the dy/dxs over

4y3ex(dy/dx)-6y5x(dy/dx)=y6-1-exy4

factor out dy/dx

(dy/dx)(4y3ex-6y5x)=y6-1-exy4

solve for dy/dx by dividing the polynomial

dy/dx=(y6-1-exy4)/(4y3ex-6y5x)

Answer 15

that makes sense, thank you so much...i will apply that same rule to all the problems :)