Question #648c3

2 Answers
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

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