How do you differentiate #r=5theta^2sectheta#?

1 Answer
Jan 18, 2017

Answer:

#(dr)/(d theta)=5thetasectheta(tantheta+2)#

Explanation:

differentiate using the #color(blue)"product rule"#

#"Given " r=f(theta).g(theta)" then"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(r'(theta)=f(theta)g'(theta)+g(theta)f'(theta))color(white)(2/2)|)))larr" product rule"#

The standard derivative of #sec(theta)# which should be known is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(d/(d theta)(sec(theta))=sec(theta)tan(theta))color(white)(2/2)|)))#

#"here " f(theta)=5theta^2rArrf'(theta)=10theta#

#"and " g(theta)=secthetarArrg'(theta)=secthetatantheta#

#rArrr'(theta)=5theta^2secthetatantheta+sectheta.10theta#

#=5thetasectheta(tantheta+2)#