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EXERCISE 1.3 · Q138

Q.Show that dydx=yx\dfrac{dy}{dx}=\dfrac{y}{x} in the following, where aa and pp are constants: log⁡x20−y20x20+y20=20\log\dfrac{x^{20}-y^{20}}{x^{20}+y^{20}}=20

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Given log⁡x20−y20x20+y20=20\log\dfrac{x^{20}-y^{20}}{x^{20}+y^{20}}=20.

Step 1 — the argument of log⁡\log is constant:

x20−y20x20+y20=k(k=e20, constant)\frac{x^{20}-y^{20}}{x^{20}+y^{20}}=k \quad(k=e^{20},\text{ constant})

Step 2 — cross-multiply:

x20−y20=k(x20+y20)x^{20}-y^{20}=k(x^{20}+y^{20})

Step 3 — differentiate implicitly:

20x19−20y19dydx=k(20x19+20y19dydx)20x^{19}-20y^{19}\frac{dy}{dx}=k\left(20x^{19}+20y^{19}\frac{dy}{dx}\right)

Step 4 — collect dy/dxdy/dx terms:

−20y19(1+k)dydx=20x19(k−1)-20y^{19}(1+k)\frac{dy}{dx}=20x^{19}(k-1)

dydx=x19(1−k)y19(1+k)\frac{dy}{dx}=\frac{x^{19}(1-k)}{y^{19}(1+k)}

Step 5 — resolve 1−k,1+k1-k,1+k from k=x20−y20x20+y20k=\dfrac{x^{20}-y^{20}}{x^{20}+y^{20}}: …

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