Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
47 views
in Calculus by (114k points)
closed by

If f(x) = x3sinx, Find f'(x)


1. (x cos x + 3sin x)
2. x2sin x)
3. x2(x cos x + 3sin x)
4. x2 cos x

1 Answer

0 votes
by (115k points)
selected by
 
Best answer
Correct Answer - Option 3 : x2(x cos x + 3sin x)

Concept:

\(\rm \frac{\mathrm{d} (x^{n})}{\mathrm{d} x} = nx^{n-1} \)

\(\rm \frac{\mathrm{d} (uv)}{\mathrm{d} x} = u\frac{\mathrm{d} v}{\mathrm{d} x} + v \frac{\mathrm{d} u}{\mathrm{d} x}\)

\(\rm \frac{\mathrm{d} (\sin x)}{\mathrm{d} x} = \cos x\)

Calculation:

Given: f(x) = x3sinx

f'(x) = \(\rm \frac{\mathrm{d} (x^{3}sinx)}{\mathrm{d} x}\)

⇒ \(\rm \frac{\mathrm{d} (x^{3}sinx)}{\mathrm{d} x} = x^{3}\frac{\mathrm{d} (sinx)}{\mathrm{d} x} + sinx \frac{\mathrm{d} x^{3}}{\mathrm{d} x}\)

= x3 cos x + sin x 3x2

= x2(x cos x + 3sin x)

∴ The required value is x2(x cos x + 3sin x).

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...