Question #648c3

2 Answers
Deepak Pandey
Sep 10, 2017

#f'(x)=6tan3xsec^2 3x#

Explanation:

We have
#f (x)=1/cos^2(3x)#
Differentiating both sides with respect to #x# using chain rule we get
#f'(x)= -2/cos^3 (3x)×(-sin3x)×3#
#=6tan3xsec^2 3x#

Jim G.
Sep 10, 2017

#6tan3xsec^2 3x#

Explanation:

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

#"given "y=(f(g(x))" then"#

#dy/dx=f'(g(x)xxg'(x)larr" chain rule"#

#y=1/(cos^2 3x)=(cos3x)^(-2)#

#rArrdy/dx-2(cos3x)^(-3)xxd/dx(cos3x)#

#color(white)(y)=-2(cos3x)^(-3)xx-sin3x xxd/dx(3x)#

#color(white)(y)=(6sin3x)/(cos3x)^3#

#color(white)(y)=6tan3x xx1/(cos 3x)^2#

#color(white)(y)=6tan3xsec^2 3x#

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