MHT CET · Maths · Differentiation
If \(y=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots \ldots\) then \(\frac{d y}{d x}=\)
- A \(y-1\)
- B \(y+1\)
- C \(y^{2}-1\)
- D \(y\)
Answer & Solution
Correct Answer
(D) \(y\)
Step-by-step Solution
Detailed explanation
Given
\(y=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots\)
\(\therefore \frac{d y}{d x}=e^{x}=y\)
\(\Rightarrow y=e^{x} \)
\(y=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots\)
\(\therefore \frac{d y}{d x}=e^{x}=y\)
\(\Rightarrow y=e^{x} \)
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