Tan(2x+3). Differentiate it?

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Answer 1 · 2018-05-12T05:30:54

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

We know that,

#color(green)((1)d/(dx)(tanx)=sec^2x#

Here,

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

Let #y=tanu ,where,color(red)( u=2x+3#

#=>(dy)/(du)=sec^2u and (du)/(dx)=2#

#"Using "color(blue)"Chain Rule:"#

#(dy)/(dx)=(dy)/(du)*(du)/(dx)#

So,

#(dy)/(dx)=sec^2uxx2=2sec^2u#

Subst. back ,#color(red)(u=2x+3#

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