LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
15 views
in Calculus by (54.3k points)
closed by

Find the differentiation of f(x) = x2.ex at x = 1


1.  2e
2. 3e
3. e
4. None of these

1 Answer

0 votes
by (30.0k points)
selected by
 
Best answer
Correct Answer - Option 2 : 3e

Concept:

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

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

\(\rm \frac{\mathrm{d} (e^{x})}{\mathrm{d} x} = e^{x}\)

Calculation:

Given: f(x) = x2.ex

Now, f'(x) = \(\rm \frac{\mathrm{d} (x^{2}e^{x})}{\mathrm{d} x}\)

By chain rule ,

⇒ \(\rm \frac{\mathrm{d} (x^{2}e^{x})}{\mathrm{d} x} = e^{x}\frac{\mathrm{d} (x^{2})}{\mathrm{d} x} + x^{2} \frac{\mathrm{d} (e^{x})}{\mathrm{d} x}\)

= ex2x +x2  ex

= xex(x + 2)

At x = 1;

f'(x)|x=1 = 3e

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

...