How do you implicitly differentiate #2=e^(xy)ln(x^3y)+siny #?
When this is done in situ it is known as implicit differentiation.
We have:
First let's simplify the expression using the rules for logarithms:
Advanced Calculus
And so:
Here we get an immediate implicit function for the derivative, so here if you are familiar with partial differentiation the solution is much easier
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To implicitly differentiate the equation ( 2 = e^{xy}  \ln(x^3y) + \sin(y) ), follow these steps:
 Differentiate each term of the equation with respect to ( x ).
 Use the chain rule and product rule where necessary.
 After differentiating, solve for ( \frac{dy}{dx} ).
Differentiating each term:

Differentiate ( e^{xy} ) using the chain rule: [ \frac{d}{dx}(e^{xy}) = ye^{xy} ]

Differentiate ( \ln(x^3y) ) using the product rule: [ \frac{d}{dx}(\ln(x^3y)) = \frac{1}{x^3y} \cdot (3x^2y + x^3 \cdot \frac{dy}{dx}) ]

Differentiate ( \sin(y) ) using the chain rule: [ \frac{d}{dx}(\sin(y)) = \cos(y) \cdot \frac{dy}{dx} ]
Now, substitute the derivatives back into the equation and solve for ( \frac{dy}{dx} ).
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To implicitly differentiate the equation (2 = e^{xy}  \ln(x^3y) + \sin(y)), we differentiate each term with respect to (x), treating (y) as a function of (x) using the chain rule.
The derivative of (2) with respect to (x) is (0) since it's a constant.
For (e^{xy}), we use the chain rule to get (e^{xy} \cdot (y + x \cdot y')).
For (\ln(x^3y)), we use the chain rule and the derivative of (\ln(u)) is (\frac{1}{u} \cdot u'), so we get (\frac{1}{x^3y} \cdot (3x^2y + x^3y')).
For (\sin(y)), the derivative is (\cos(y) \cdot y').
Putting it all together, we have:
[ 0 = e^{xy} \cdot (y + x \cdot y')  \frac{1}{x^3y} \cdot (3x^2y + x^3y') + \cos(y) \cdot y' ]
Simplifying gives:
[ 0 = e^{xy} \cdot (y + x \cdot y')  \frac{3}{x^2}  \frac{1}{y} + \cos(y) \cdot y' ]
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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