How do you implicitly differentiate #-y=xy-xe^-y #?
As a Second Method , we can solve the Problem without the
use of Implicit Diffn.
I leave it, as an exercise, to show that our both Answers match!
By signing up, you agree to our Terms of Service and Privacy Policy
Since, the Problem is to be solved using Implicit Diffn., we will
first solve it accordingly.
By the Product Rule , then,
Enjoy Maths.!
By signing up, you agree to our Terms of Service and Privacy Policy
To implicitly differentiate the equation ( -y = xy - xe^{-y} ), follow these steps:
- Differentiate both sides of the equation with respect to ( x ).
- Apply the product rule when differentiating terms involving ( y ).
- Use the chain rule when differentiating ( e^{-y} ).
Differentiating ( -y ) with respect to ( x ) gives ( -\frac{dy}{dx} ).
For the term ( xy ), apply the product rule: ( \frac{d(xy)}{dx} = x\frac{dy}{dx} + y ).
For the term ( xe^{-y} ), differentiate ( x ) with respect to ( x ) (which gives 1), then differentiate ( e^{-y} ) using the chain rule: ( \frac{d(e^{-y})}{dx} = -e^{-y} \frac{dy}{dx} ). So, ( \frac{d(xe^{-y})}{dx} = 1 \cdot e^{-y} + x \cdot (-e^{-y}) \frac{dy}{dx} ).
Combine these results to get:
[ -\frac{dy}{dx} = x\frac{dy}{dx} + y - e^{-y} + x(-e^{-y})\frac{dy}{dx} ]
Rearrange the terms and solve for ( \frac{dy}{dx} ):
[ -\frac{dy}{dx} - x\frac{dy}{dx} - x(-e^{-y})\frac{dy}{dx} = y - e^{-y} ]
[ (-1 - x - xe^{-y})\frac{dy}{dx} = y - e^{-y} ]
[ \frac{dy}{dx} = \frac{y - e^{-y}}{-1 - x - xe^{-y}} ]
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
- What is the slope of the tangent line of # x/(x-3y) = C #, where C is an arbitrary constant, at #(-2,1)#?
- How do you differentiate #f(x) = sqrt(arctan(3x) # using the chain rule?
- How do you use the chain rule to differentiate #y=sqrt(x^2-7x)#?
- What is the slope of the tangent line of # (x-y^2)/(xe^(x-y^2)) =C #, where C is an arbitrary constant, at #(1,1)#?
- How do you find the derivative of #f(t)=-2t^2+3t-6#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7