MHT CET · Maths · Differentiation
If \(x=\log t, y+1=\frac{1}{t}, \quad\) then \(e^{-x} \frac{d^{2} x}{d y^{2}}+\frac{d x}{d y}=\)
- A 0
- B 2
- C \(-1\)
- D 1
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
Given \(x=\log t\) and \(y+1=\frac{1}{t}\)
\(\therefore \frac{d x}{d t}=\frac{1}{t}\) and \(\frac{d y}{d t}=\frac{-1}{t^{2}}\)
\(\therefore \frac{d x}{d y}=\frac{1}{t} \times\left(-t^{2}\right)=-t\)
\(\therefore \frac{d^{2} x}{d y^{2}}=\frac{d}{d t}\left(\frac{d x}{d y}\right) \times \frac{d t}{d y}=\frac{d}{d t}(-t) \times \frac{1}{\left(\frac{d y}{d t}\right)}=\frac{(-1)}{\left(\frac{-1}{t^{2}}\right)}=t^{2}\)
\(e^{-x}=e^{-\log t}=e^{\log (t)^{-1}}=\frac{1}{t}\)
Thus \(e^{-x} \frac{d^{2} y}{d x^{2}}+\frac{d x}{d y}\)
\(=\left(\frac{1}{t}\right)\left(t^{2}\right)+(-t)=t-t=0\)
\(\therefore \frac{d x}{d t}=\frac{1}{t}\) and \(\frac{d y}{d t}=\frac{-1}{t^{2}}\)
\(\therefore \frac{d x}{d y}=\frac{1}{t} \times\left(-t^{2}\right)=-t\)
\(\therefore \frac{d^{2} x}{d y^{2}}=\frac{d}{d t}\left(\frac{d x}{d y}\right) \times \frac{d t}{d y}=\frac{d}{d t}(-t) \times \frac{1}{\left(\frac{d y}{d t}\right)}=\frac{(-1)}{\left(\frac{-1}{t^{2}}\right)}=t^{2}\)
\(e^{-x}=e^{-\log t}=e^{\log (t)^{-1}}=\frac{1}{t}\)
Thus \(e^{-x} \frac{d^{2} y}{d x^{2}}+\frac{d x}{d y}\)
\(=\left(\frac{1}{t}\right)\left(t^{2}\right)+(-t)=t-t=0\)
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