Differentiate d/dx(x3.sin4x)
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by M G 8 years, 8 months ago
- 1 answers
Test
Related Questions
Posted by M D 1 year, 4 months ago
- 1 answers
Posted by Pankaj Tripathi 1 year, 5 months ago
- 1 answers
Posted by Mohammed Javith 1 year, 5 months ago
- 0 answers
Posted by Nekita Baraily 1 year, 4 months ago
- 2 answers
Posted by Mansi Class 9Th 1 year, 5 months ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Payal Singh 8 years, 8 months ago
Use chain rule for it,
{tex}{d\over dx}(x^3.sin4x){/tex}
{tex}= x^3.{d\over dx}(sin 4x) + sin4x. {d\over dx} (x^3){/tex}
{tex}= x^3.cos4x.{d\over dx}(4x) + sin4x.3x^2{/tex}
{tex}= 4x^3cos4x + 3x^2sin4x{/tex}
1Thank You