If f(x+ay)+g(x-ay)=0,then a frac{dy}{dx}= | vedclass

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(B) Given the equation: $f(x+ay)+g(x-ay)=0$. Differentiating both sides with respect to $x$ using the chain rule: $f^{\prime}(x+ay)(1+a \frac{dy}{dx})

Vedclass · Accepted Answer

$\frac{f^{\prime}(x+ay)+g^{\prime}(x-ay)}{g^{\prime}(x-ay)-f^{\prime}(x+ay)}$

Vedclass · Answer

(B) Given the equation: $f(x+ay)+g(x-ay)=0$. Differentiating both sides with respect to $x$ using the chain rule: $f^{\prime}(x+ay)(1+a \frac{dy}{dx}) + g^{\prime}(x-ay)(1-a \frac{dy}{dx}) = 0$. Expanding the terms: $f^{\prime}(x+ay) + a \frac{dy}{dx} f^{\prime}(x+ay) + g^{\prime}(x-ay) - a \frac{dy}{dx} g^{\prime}(x-ay) = 0$. Rearranging to isolate the terms with $\frac{dy}{dx}$: $a \frac{dy}{dx} [f^{\prime}(x+ay) - g^{\prime}(x-ay)] = -[f^{\prime}(x+ay) + g^{\prime}(x-ay)]$. Therefore: $a \frac{dy}{dx} = \frac{-(f^{\prime}(x+ay) + g^{\prime}(x-ay))}{f^{\prime}(x+ay) - g^{\prime}(x-ay)} = \frac{f^{\prime}(x+ay) + g^{\prime}(x-ay)}{g^{\prime}(x-ay) - f^{\prime}(x+ay)}$.

If $f(x+ay)+g(x-ay)=0$,then $a \frac{dy}{dx}=$

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