# Implicit differentiation calculus

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find dy/dx by implicit differentiation

a) xy + x^5y^2 + 3x^3 - 4 = 1

b) lny = cos x

thank you

• Implicit differentiation calculus -

normally you just have an equation like

y = 3x^2
and you take the derivative of both sides, to get
y' = 6x

Then you advance to using the chain rule and product rule, and you have something like

y = e^x*sin^2(6x)
and you take the derivative of both sides to get
y' = e^x(sin^2(6x)+12sin(6x)cos(6x))

implicit differentiation just uses these techniques. Take the derivative of both sides of the equation, using the chain rule, and remembering that dx/dx = x' = 1 and dy/dx = y'

a) the derivative of xy is

d/dx(xy) = x'y + xy' = y + xy'
d/dx(x^5y^2) = (5x^4)(y^2) + (x^5)(2yy')
d/dx(3x^3) = 9x^2
d/dx(-4) = 0
d/dx(1) = 0
and put it all together to get
y + xy' + 5x^4y^2 + 2x^5yy' + 9x^2 = 0
Now just solve for y' by collecting terms
y'(x+2x^5y) + (y + 5x^4y^2 + 9x^2) = 0

y' = -(y + 5x^4y^2 + 9x^2)/(x+2x^5y)

b) again, use the chain rule:
d/dx(ln y) = d/dy(ln y)*dy/dx = 1/y y'
1/y y' = -sin x
y' = -y sin x

• Implicit differentiation calculus -

thank for you so much steve!!!

i just wanted to double check.. for (a) it is y' = - (y+5x^4y^2+9x^2)/(x+2x^5y)

i was wondering if the last part of it (x+2x^5y) is it 2x^(5y) or 2x^5(y). im not sure if the "y" is in the exponent or not.

• Implicit differentiation calculus -

No, since we took the derivative of x^5*y^2 we get (5x^4)(y^2) + (x^5)(2yy')

That is, 2(x^5)(y)

• Implicit differentiation calculus -

ok so the final answer would b

a) y' = -(y + 5x^4y^2 + 9x^2)/(x+2x^5y)

and

b) y' = -y sin x

thank you SO much once again steve, my exam is in a couple of hours and this solution really helped me, im very thankful. thanks so much!!

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