How do you find #(dy)/(dx)# given #xy=tanxy#?
now
#d/(dx)f(y(x),x) = y(x) + x y(x)  Sec(x y(x))^2 (y(x) + x y'(x)) = 0#
We can also obtain the same result straightforward
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To find ( \frac{dy}{dx} ) given ( xy = \tan(xy) ), you can use implicit differentiation.
 Differentiate both sides of the equation with respect to (x).
 Apply the product rule and chain rule where necessary.
 Solve for ( \frac{dy}{dx} ).
The steps are as follows:

Start with ( xy = \tan(xy) ).

Differentiate both sides with respect to (x):
[ \frac{d}{dx}(xy) = \frac{d}{dx}(\tan(xy)) ]

Apply the product rule and chain rule:
[ y + x\frac{dy}{dx} = \sec^2(xy)\left(y + x\frac{dy}{dx}\right) ]

Solve for ( \frac{dy}{dx} ):
[ y + x\frac{dy}{dx} = \sec^2(xy)y + x\sec^2(xy)\frac{dy}{dx} ] [ x\frac{dy}{dx}  x\sec^2(xy)\frac{dy}{dx} = \sec^2(xy)y  y ] [ x\frac{dy}{dx}(1  \sec^2(xy)) = \sec^2(xy)y  y ] [ x\frac{dy}{dx}(\tan^2(xy)) = \sec^2(xy)y  y ] [ \frac{dy}{dx} = \frac{\sec^2(xy)y  y}{x\tan^2(xy)} ]
So, ( \frac{dy}{dx} = \frac{\sec^2(xy)y  y}{x\tan^2(xy)} ).
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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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