Also I'm gonna need solutions please

in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

(1) If you can’t remember the derivative of arcsec, here’s how you can do this:

sec(y)=√x, sec²(y)=x, 1+tan²(y)≡sec²y, tan(y)=√(sec²(y)-1)=√(x-1);

sec(y)tan(y)y'=1/(2√x),

(√x)(√(x-1))y'=1/(2√x),

y'=dy/dx=1/(2x√(x-1)), which can also be written ½√(x-1)/(x²-x).

(2) Let y=ux, dy/dx=y'=u+u'x;

y/x=arctan(4x), u=arctan(4x), tan(u)=4x, sec²(u)u'=4, (1+tan²(u))u'=4, (1+16x²)u'=4, u'=4/(1+16x²);

y'=u+u'x=arctan(4x)+4x/(1+16x²).

(3) y=arccsc(x³), csc(y)=x³, sin(y)=1/x³, cos(y)=√(1-1/x⁶)=√(x⁶-1)/x³;

y'cos(y)=-3/x⁴, y'√(x⁶-1)/x³=-3/x⁴, y'√(x⁶-1)=-3/x, y'=-3/(x√(x⁶-1)) which can be written -3√(x⁶-1)/(x⁷-x).

(4) y=arccot(1-2x), cot(y)=1-2x, tan(y)=1/(1-2x), tan²(y)=1/(1-2x)², sec²(y)=1+1/(1-2x)²=(2-4x+4x²)/(1-2x)²=2(1-2x+2x²)/(1-2x)²;

y'sec²(y)=2/(1-2x)², 2(1-2x+2x²)/(1-2x)²y'=2/(1-2x)², y'=1/(1-2x+2x²).

by Top Rated User (1.1m points)

Related questions

1 answer
1 answer
1 answer
1 answer
asked Apr 24, 2013 in Calculus Answers by anonymous | 964 views
2 answers
asked Nov 12, 2011 in Trigonometry Answers by anonymous | 1.9k views
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
87,540 questions
99,812 answers
2,417 comments
523,783 users