Answer to Question #118866 in Calculus for Nimra

Question #118866
f³f¹(x) = √secx.x³
1
Expert's answer
2020-06-01T18:22:33-0400
f(x)=secxx3f(x)=\sqrt{\sec x}\cdot x^3

dfdx=x32secx(sinxcos2x)+3x2secx={df\over dx}={x^3\over 2\sqrt{\sec x}}(-{-\sin x\over \cos^2x})+3x^2\sqrt{\sec x}=

=x2xtanx+62cosxsecx=x2xtanx+62cosx==x^2{x\tan x+6\over 2\cos x\sqrt{\sec x}}=x^2{x\tan x+6\over 2\sqrt{\cos x}}=

=x2secxxtanx+62=x^2\sqrt{\sec x}\cdot{x\tan x+6\over 2}




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment