Question
Download Solution PDF\(\rm d\left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!} \ldots\right) \)/dx is equal to:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The given function is a power series:
\[ f(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \]
This is the Taylor (Maclaurin) series expansion of \(\sin(x)\).
Calculation:
Differentiate both sides with respect to x:
\[ \frac{d}{dx} \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right) = 1 - \frac{3x^2}{3!} + \frac{5x^4}{5!} - \frac{7x^6}{7!} + \cdots \]
This new series is the Maclaurin expansion of \(\cos(x)\).
Last updated on Jun 24, 2025
-> ISRO Scientist Engineering apply online 2025 link has been activated (ISRO:ICRB:03(CEPO):2025).
-> A total of 39 vacancies are announced for the ISRO recruitment in Civil, RAC and other disciplines
-> ISRO Scientist Engineering recruitment 2025 notification has been released for 320 vacancies. The last date to fill ISRO application form is June 16 (Advt No. ISRO:ICRB:02(EMC):2025).
-> ISRO Scientist Engineer recruitment 2025 for 31 vacancies has been released.
->ISRO Scientist recruitment 2025 notification has been released.
->The last date to apply for ISRO scientist recruitment 2025 is May 30 ( Advt.No. ISRO:ICRB:01:(EMC):2025).
->Candidates with BE/BTech degree in the respective discipline can only apply for ISRO Scientist recruitment 2025.
-> Candidates can refer ISRO Scientist previous year paper to prepare for the exam.