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Differentiation — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferentiation5 Marks
Question
If $\text{x}=\Big(\text{t}+\frac{1}{\text{t}}\Big)^\text{a},\text{y}=\text{a}^{\text{t}+\frac{1}{\text{t}}},$ find $\frac{\text{dy}}{\text{dx}}$
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Answer

Here, $\text{x}=\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a}}$
Differentiating it with respect to t using chain rule,
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\left[\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a}}\right]$
$=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a-1}}\frac{\text{d}}{\text{dt}}\Big(\text{t}+\frac{1}{\text{t}}\Big)$
$\frac{\text{dx}}{\text{dt}}=\text{a}\Big(\text{t}+\frac{1}{\text{t}}\Big)^{\text{a-1}}\Big(\text{1}-\frac{1}{\text{t}^{2}}\Big)\ .....(\text{i})$
And, $\text{y}=\text{a}^{(\text{t}+\frac{1}{\text{t}})}$
Differentiating it with respect to t using chain rule,
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\bigg[\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\bigg]$
$=\frac{\text{d}}{\text{dt}}\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\frac{\text{d}}{\text{dt}}\left(\text{t}+\frac{1}{\text{t}}\right)$
$\frac{\text{dy}}{\text{dt}}=\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\text{a}\Big(\text{t}-\frac{1}{\text{t}^{2}}\Big)\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\text{a}\left(1-\frac{1}{\text{t}^{2}}\right)}{\text{a}\big(\text{t}+\frac{1}{\text{t}}\big)^{\text{a}-1}\big(1-\frac{1}{\text{t}}^{2}\big)}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{a}^{\big(\text{t}+\frac{1}{\text{t}}\big)}\times\log\text{a}}{\text{a}\big(\text{t}+\frac{1}{\text{t}}\big)^{\text{a}-1}}$

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