Question

if t=√y+1 find dy/dt​

Answer 1

Answer:

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Step-by-step explanation:

Given :

$\displaystyle \sf t=\sqrt{y}+1$

Rewrite as :

$\displaystyle \sf \longrightarrow t-1=\sqrt{y}$

$\displaystyle \sf \longrightarrow y= (t-1)^2$

Now diff. w.r.t. t we get :

$\displaystyle \sf \longrightarrow \frac{dy}{dt} =\frac{d}{dt} (t-1)^2$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)^{2-1}.(t-1)'$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(t'-1')$

We know derivative of constant is zero!

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(1-0)$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Hence we get required answer!

Answer 2

Answer:

$Answer:</p><p></p><p>\begin{gathered}\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1) \\\end{gathered}⟶dtdy=2.(t−1)</p><p></p><p>Step-by-step explanation:</p><p></p><p>Given :</p><p></p><p>\begin{gathered}\displaystyle \sf t=\sqrt{y}+1 \\ \\\end{gathered}t=y+1</p><p></p><p>Rewrite as :</p><p></p><p>\begin{gathered}\displaystyle \sf \longrightarrow t-1=\sqrt{y} \\ \\\end{gathered}⟶t−1=y</p><p></p><p>\begin{gathered}\displaystyle \sf \longrightarrow y= (t-1)^2 \\ \\\end{gathered}⟶y=(t−1)2</p><p></p><p>Now diff. w.r.t. t we get :</p><p></p><p>\begin{gathered}\displaystyle \sf \longrightarrow \frac{dy}{dt} =\frac{d}{dt} (t-1)^2 \\ \\\end{gathered}⟶dtdy=dtd(t−1)2</p><p></p><p>\begin{gathered}\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)^{2-1}.(t-1)' \\ \\\end{gathered}⟶dtdy=2.(t−1)2−1.(t−1)′</p><p></p><p>\begin{gathered}\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(t'-1') \\ \\\end{gathered}⟶dtdy=2.(t−1).(t′−1′)</p><p></p><p>We know derivative of constant is zero!</p><p></p><p>\begin{gathered}\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(1-0) \\ \\\end{gathered}⟶dtdy=2.(t−1).(1−0)</p><p></p><p>\begin{gathered}\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1) \\ \\\end{gathered}⟶dtdy=2.(t−1)</p><p></p><p>Hence we get required answer!</p><p></p><p>$