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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation4 Marks
Question
If $x=f(t)$ and $y=g(t)$ are differentiable function of $t$ so that $y$ is a differentiable function of $x$ and $\frac{d x}{d t} \neq 0$, then prove that :
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Hence find $\frac{d y}{d x}$ if $x=\sin t$ and $y=\operatorname{cost} t$.
✓

Answer

Given: $x=f(t), y=g(t)$
Let $\delta t$ be a small increment in $t, \delta x, \delta y$ be. the corresponding increments in $x$ and $y$ respectively.
Consider
$\frac{\delta y}{\delta x}=\frac{\frac{\delta y}{\delta t}}{\frac{\delta x}{\delta t}}, \frac{\delta x}{\delta t} \neq 0$
Taking the limit as $\delta t \rightarrow 0$,
$\lim _{x \rightarrow 00} \frac{\delta y}{\delta x}=\lim _{x t \rightarrow 0} \frac{\delta y}{\frac{\delta t}{\delta x}}, \lim _{x \rightarrow 00} \frac{\delta x}{\delta t} \neq 0$
As $\delta t \rightarrow 0, \delta x \rightarrow 0$
$\therefore \lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta t}}{\lim _{\delta x \rightarrow 0} \frac{\delta x}{\delta t}}....(1)$
As $x$ and $y$ are differentiable function of $t$
$\lim _{\delta \rightarrow 00} \frac{\delta x}{\delta t}=\frac{d x}{d t} \text { and } \lim _{\delta t \rightarrow 0} \frac{\delta y}{\delta t}=\frac{d y}{d t}....(2)$
As $\text{R.H.S.}$ limit on $(1)$ exists and are finite $\text{L.H.S.}$ limit also exists and is finite $(2)$.
$\therefore \lim _{d x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}....(3)$
But $\lim _{d x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Now, given
$x=\sin t, y=\cos t$
Differentiating w.r.t. $t$
$\therefore \frac{d x}{d t}=\cos t, \frac{d y}{d t}=-\sin t$
$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-\sin t}{\cos t}=-\tan t$

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