How do you implicitly differentiate #2= ysinx-xcosy #?
Start with the given equation and differentiate with respect to x both sides of the equation
by symmetric property of equality we have
God bless....I hope the explanation is useful.
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To implicitly differentiate (2 = y \sin(x) - x \cos(y)):
- Differentiate each term of the equation with respect to (x).
- Apply the chain rule where necessary.
- Solve for (\frac{{dy}}{{dx}}) to find the derivative of (y) with respect to (x).
Differentiating (2 = y \sin(x) - x \cos(y)) with respect to (x):
[ \frac{{d}}{{dx}}(2) = \frac{{d}}{{dx}}(y \sin(x)) - \frac{{d}}{{dx}}(x \cos(y)) ]
Simplify the derivatives:
[ 0 = y\cos(x) + \sin(x)\frac{{dy}}{{dx}} + \cos(y) - x\left(-\sin(y)\frac{{dy}}{{dx}}\right) ]
Rearrange the terms to isolate (\frac{{dy}}{{dx}}):
[ \sin(x)\frac{{dy}}{{dx}} + \sin(y)\frac{{dy}}{{dx}} = -y\cos(x) - \cos(y) + x\sin(y) ]
Factor out (\frac{{dy}}{{dx}}):
[ \frac{{dy}}{{dx}}(\sin(x) + \sin(y)) = -y\cos(x) - \cos(y) + x\sin(y) ]
Finally, solve for (\frac{{dy}}{{dx}}):
[ \frac{{dy}}{{dx}} = \frac{{-y\cos(x) - \cos(y) + x\sin(y)}}{{\sin(x) + \sin(y)}} ]
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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.
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