NCERT Class 12 Mathematics — Continuity and Differentiability, Exercise 5.1, Question 10.
Problem
For a positive constant $a$, find $\dfrac{dy}{dx}$, where $y=a^{t+\frac1t}$ and $x=\left(t+\dfrac1t\right)^a$.
Key insight. Both $y$ and $x$ are functions of the same combined expression $t+\tfrac1t$, not of $t$ directly in any simpler way. Naming that combination $u$ turns this into an ordinary parametric-differentiation problem in $u$ — and since $\dfrac{dy}{dx}=\dfrac{dy/du}{dx/du}$, whatever $\dfrac{du}{dt}$ actually is never even needs to be computed, because it cancels.
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Approach
Substitute $u=t+\dfrac1t$, so that $y=a^u$ and $x=u^a$ are both plain functions of $u$ alone. Differentiate each with respect to $u$, then divide to get $\dfrac{dy}{dx}$ — the chain-rule factor $\dfrac{du}{dt}$ that would otherwise appear in both $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ cancels automatically in the ratio.
Solution
Step 1 — Introduce u = t + 1/t
$$y = a^u, \qquad x = u^a$$
Step 2 — Differentiate y with respect to u
$$\frac{dy}{du} = a^u\ln a$$
Step 3 — Differentiate x with respect to u
$$\frac{dx}{du} = a\cdot u^{a-1}$$
Step 4 — Form dy/dx as the ratio
Since $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{(dy/du)(du/dt)}{(dx/du)(du/dt)}=\dfrac{dy/du}{dx/du}$, the $du/dt$ factor cancels:
$$\frac{dy}{dx} = \frac{a^u\ln a}{a\cdot u^{a-1}}$$
Step 5 — Substitute back u = t + 1/t
$$\frac{dy}{dx} = \frac{a^{t+\frac1t}\ln a}{a\left(t+\dfrac1t\right)^{a-1}}$$
Answer
$$\frac{dy}{dx} = \frac{a^{t+\frac1t}\ln a}{a\left(t+\dfrac1t\right)^{a-1}}$$
Common mistakes
- Differentiating $y$ and $x$ with respect to $t$ directly, term by term, before spotting the shared expression $u=t+\tfrac1t$. This works but forces carrying around the extra factor $\dfrac{du}{dt}=1-\dfrac1{t^2}$ in both derivatives — extra algebra that turns out to cancel anyway once the ratio $dy/dx$ is formed.
- Treating $a^u$ like a polynomial and differentiating it as $u\cdot a^{u-1}$. Since $a$ is the base and $u$ is the exponent (not the other way around), the correct derivative uses $\dfrac{d}{du}(a^u)=a^u\ln a$, not the power rule.
Practise next
- For $y=a^{t^2+1}$ and $x=\left(t^2+1\right)^a$, find $\dfrac{dy}{dx}$, using the same substitution approach with $u=t^2+1$.
Show answer
$\dfrac{dy}{dx}=\dfrac{a^{t^2+1}\ln a}{a\left(t^2+1\right)^{a-1}}$. With $u=t^2+1$, both are functions of $u$: $y=a^{u}$ and $x=u^{a}$.
So $\dfrac{dy}{du}=a^{u}\ln a$ and $\dfrac{dx}{du}=a\,u^{a-1}$ — one is an exponential in $u$, the other a power of $u$, and they differentiate differently.
Dividing, $\dfrac{dy}{dx}=\dfrac{a^{u}\ln a}{a\,u^{a-1}}$, then put $u=t^2+1$ back. The factor $2t$ from $\dfrac{du}{dt}$ would appear in both derivatives and cancel, which is why going through $u$ directly is cleaner.
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