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EXERCISE 1.4 · Q156

Q.Find dydx\dfrac{dy}{dx} if x=(t+1t)ax=\left(t+\dfrac1t\right)^a, y=at+1ty=at+\dfrac1t, where a>0, a≠1, t≠0a>0,\ a\ne1,\ t\ne0.

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We have x=(t+1t)ax=\left(t+\dfrac1t\right)^a and y=at+1ty=at+\dfrac1t.

Step 1. Differentiate xx using the chain/power rule (exponent aa is a plain constant here):

dxdt=a(t+1t)a−1(1−1t2)\frac{dx}{dt}=a\left(t+\frac1t\right)^{a-1}\left(1-\frac1{t^2}\right)

Step 2. Differentiate yy:

dydt=a−1t2\frac{dy}{dt}=a-\frac1{t^2}

Step 3. Form the ratio:

dydx=a−1t2a(t+1t)a−1(1−1t2)\frac{dy}{dx}=\frac{a-\dfrac1{t^2}}{a\left(t+\dfrac1t\right)^{a-1}\left(1-\dfrac1{t^2}\right)} …

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