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Miscellaneous Examples · Example 42

Q.For a positive constant aa, find dydx\frac{dy}{dx}, where y=at+1ty = a^{t + \frac{1}{t}}, and x=(t+1t)ax = \left(t + \frac{1}{t}\right)^a.

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Differentiating yy and xx with respect to tt and taking the ratio, the common factor (1−1t2)\left(1-\frac1{t^2}\right) cancels, giving dydx=a t+1/tlog⁡aa(t+1t)a−1\dfrac{dy}{dx}=\dfrac{a^{\,t+1/t}\log a}{a\left(t+\frac1t\right)^{a-1}}.

When xx and yy are both given through a parameter tt, we don't eliminate tt; we differentiate each with respect to tt and use

dydx=dy/dtdx/dt,dxdt≠0.\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\qquad \frac{dx}{dt}\neq 0.

Let u=t+1tu = t+\dfrac1t, so dudt=1−1t2\dfrac{du}{dt}=1-\dfrac1{t^2}.

1. Differentiate y=auy=a^{u}

aa is a positive constant, so ddtau=aulog⁡a⋅dudt\frac{d}{dt}a^{u}=a^{u}\log a\cdot\frac{du}{dt}:

dydt=at+1/tlog⁡a(1−1t2).\frac{dy}{dt}=a^{t+1/t}\log a\left(1-\frac1{t^2}\right).

2. Differentiate x=uax=u^{a}

This is a power with the constant exponent aa, so ddtua=a ua−1⋅dudt\frac{d}{dt}u^{a}=a\,u^{a-1}\cdot\frac{du}{dt}:

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

3. Take the ratio

dydx=at+1/tlog⁡a(1−1t2)a(t+1t)a−1(1−1t2).\frac{dy}{dx}=\frac{a^{t+1/t}\log a\left(1-\frac1{t^2}\right)}{a\left(t+\frac1t\right)^{a-1}\left(1-\frac1{t^2}\right)}. …

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