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Chapter 6: Problem 62

If \(\log (x+y)=2 x y\), then prove that \(y^{\prime}(0)=1\).

Short Answer

Expert verified
After performing implicit differentiation on the given equation, isolating \(y'\) and substituting \(x = 0\), it is indeed proved that \(y'(0) = 1\).

Step by step solution

01

Apply Implicit Differentiation

Apply the derivative to both sides of the equation \(\log(x+y)=2 x y\). Using the chain rule, the derivative of \(\log(x + y)\) is \(\frac{1}{x+y}\) times the derivative of the inner function which is \(1 + y'\), because the derivative of \(y\) with respect to \(x\) is \(y'\). On the other side, the derivative of \(2 x y\) can be computed using the product rule to be \(2 y + 2 x y'\). Therefore, the resulting equation after taking derivatives is \(\frac{1+ y'}{x+y} = 2y + 2x y'\).
02

Isolate y'

The next step is to isolate \(y'\) in the equation. Multiply each side of the equation by \(x + y\) to get \(y' + 1 = 2 y (x + y) + 2 x y (x + y)\). Now isolate \(y'\) by grouping terms with \(y'\) on one side of the equation and the remaining terms on the other side. This results in \(y' = 2 y (x + y) + 2xy(x + y) - 1\).
03

Substitute x=0

Now we must substitute \(x=0\) into the equation for \(y'\). Doing so simplifies the equation to \(y'(0) = 2 y(0) - 1\). As per the given condition, when \(x=0\), \(y=1\) (This is inferred from the equation \(\log(x+y)=2 x y\), where x is 0, we get \(y=1\)). Therefore, we can substitute \(y(0) = 1\) into the equation, yielding \(y'(0) = 2*1 - 1 = 1\).
04

Conclude with proof

The derivative at point \(0\) evaluated to \(1\), thus \(y^{\prime}(0)=1\) is proven to be true.

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