Question #9409e

2 Answers
Jan 9, 2018

#4secxtanx+2cscxcotx#

Explanation:

#"differentiate using "color(blue)"standard derivatives"#

#•color(white)(x)d/dx(secx)=secxtanx#

#•color(white)(x)d/dx(cscx)=-cscxcotx#

#rArrd/dx(4secx-2cscx)#

#=4secxtanx-2(-cscxcotx)#

#=4secxtanx+2cscxcotx#

Jan 9, 2018

#d/dx(4sec(x)-2csc(x))=4sec(x)tan(x)+2csc(x)cot(x)#

Explanation:

We have your original question

#d/dx(4sec(x)-2csc(x))#

Thankfully, this question isn't too bad, since the only operation between the x values is subtraction. All we really need to know for this question is

#d/dx(sec(x))=sec(x)tan(x)#

and

#d/dx(csc(x))=-csc(x)cot(x)#

So we can simplify your original question to

#d/dx4sec(x)- d/dx2csc(x)#

Since there are constants, we can take those out.

#4d/dxsec(x)- 2d/dxcsc(x)#

Finally, we can substitute what we found earlier to get the following:

#4sec(x)tan(x)-2(-csc(x)cot(x))#

And since there's a -2 being multiplied to -csc(x), the subtractions cancel out to get your final answer.

#4sec(x)tan(x)+2csc(x)cot(x)#