How do you implicitly differentiate #2=e^(xy)-xcosy #?
Start from the given equation
differentiate both sides of the equation with respect to x:
solving now for first derivative y'
final answer becomes
By signing up, you agree to our Terms of Service and Privacy Policy
To implicitly differentiate ( 2 = e^{xy} - x \cos(y) ), we follow these steps:
- Differentiate both sides of the equation with respect to ( x ).
- Apply the chain rule and product rule where necessary.
- Solve for ( \frac{{dy}}{{dx}} ).
Starting with the given equation:
[ 2 = e^{xy} - x \cos(y) ]
Differentiating both sides with respect to ( x ):
[ \frac{{d}}{{dx}} (2) = \frac{{d}}{{dx}} (e^{xy} - x \cos(y)) ]
[ 0 = \frac{{d}}{{dx}} (e^{xy}) - \frac{{d}}{{dx}} (x \cos(y)) ]
Using the chain rule for the first term and product rule for the second term:
[ 0 = e^{xy} \left( \frac{{d}}{{dx}} (xy) \right) - \left( \frac{{d}}{{dx}} (x) \cdot \cos(y) + x \cdot \frac{{d}}{{dx}} (\cos(y)) \right) ]
[ 0 = e^{xy} \left( y + x \frac{{dy}}{{dx}} \right) - (1 \cdot \cos(y) + x \cdot (-\sin(y) \cdot \frac{{dy}}{{dx}})) ]
Now, isolate ( \frac{{dy}}{{dx}} ):
[ e^{xy} y + e^{xy}x \frac{{dy}}{{dx}} = \cos(y) + x \sin(y) \frac{{dy}}{{dx}} ]
[ e^{xy}x \frac{{dy}}{{dx}} - x \sin(y) \frac{{dy}}{{dx}} = \cos(y) - e^{xy} y ]
[ \frac{{dy}}{{dx}} (e^{xy}x - x \sin(y)) = \cos(y) - e^{xy} y ]
[ \frac{{dy}}{{dx}} = \frac{{\cos(y) - e^{xy} y}}{{e^{xy}x - x \sin(y)}} ]
By signing up, you agree to our Terms of Service and Privacy Policy
To implicitly differentiate (2 = e^{xy} - x \cos y), differentiate each term with respect to (x) using the chain rule for (e^{xy}) and the product rule for (xy).
[\frac{d}{dx}(2) = \frac{d}{dx}(e^{xy}) - \frac{d}{dx}(x\cos y)]
[0 = e^{xy} \frac{d}{dx}(xy) - (\cos y + x\frac{d}{dx}(\cos y))]
[0 = e^{xy}(y + x\frac{dy}{dx}) - (\cos y - x\sin y\frac{dy}{dx})]
[0 = ye^{xy} + x^2\frac{dy}{dx}e^{xy} - \cos y + x\sin y\frac{dy}{dx}]
Solve for (\frac{dy}{dx}).
[\frac{dy}{dx}(x^2e^{xy} + x\sin y) = \cos y - ye^{xy}]
[\frac{dy}{dx} = \frac{\cos y - ye^{xy}}{x^2e^{xy} + x\sin 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.
- How do you differentiate #f(x)=(3x-cos^3x)^2/4# using the chain rule?
- How do you differentiate #y = (2x^4 - 3x) / (4x - 1)#?
- How do you differentiate #y=5x^3(sinx)^2#?
- How do you find #dy/dx# by implicit differentiation of #x^3+y^3=4xy+1# and evaluate at point (2,1)?
- How do you implicitly differentiate #11=(x-y)/(e^y-e^x)#?
- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7