How do you differentiate #f(x)=2sinx-tanx#?

2 Answers
Apr 30, 2018

The derivative is #2Cos(x)-(1/Cos^2(x))#- see below for how to do it.

Explanation:

If

#f(x)=2Sinx-Tan(x)#

For the sine part of the function, the derivative is simply: #2Cos(x)#

However, #Tan(x)# is a bit more tricky- you have to use the quotient rule.

Recall that #Tan(x)=(Sin(x)/Cos(x))#

Hence we can use The quotient rule

if#f(x)=(Sin(x)/Cos(x))#

Then

#f'(x)=((Cos^2(x)-(-Sin^2(x)))/(Cos^2(x)))#

#Sin^2(x)+Cos^2(x)=1#

#f'(x)=1/(Cos^2(x))#

So the complete function becomes
#f'(x)=2Cos(x)- (1/Cos^2(x))#

Or

#f'(x)=2Cos(x)-Sec^2(x)#

Apr 30, 2018

#f'(x)=2cosx-sec^2x#

Explanation:

#"utilising the "color(blue)"standard derivatives"#

#•color(white)(x)d/dx(sinx)=cosx" and "d/dx(tanx)=sec^2x#

#rArrf'(x)=2cosx-sec^2x#