Question #9c03c

2 Answers
James W.
Dec 15, 2017

#sec^2(2x+3) * 2#

Explanation:

#f(x) = tan(2x+3)#

The chain rule states that:

#(df(u))/dx = (df)/(du) * (du)/(dx)#

Setting #u = 2x+3#, we get the equation:

#(df(x))/dx = d/(du) tan(u) * d/(dx) 2x+3#

And since #d/(du) tan(u)# = #sec^2(u)# and #d/(dx) 2x+3# = 2:

#(df(x))/dx = sec^2(u) * 2#, which is:

# = sec^2(2x+3) * 2#

Jim G.
Dec 15, 2017

#dy/dx=2sec^2(2x+3)#

Explanation:

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

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

#dy/dx=f'(g(x))xxg'(x)#

#y=tan(2x+3)#

#rArrdy/dx=sec^2(2x+3)xxd/dx(2x+3)#

#color(white)(rArrdy/dx)=2sec^2(2x+3)#

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