How do you differentiate #f(x)=sqrttan(2-x^3) # using the chain rule?

2 Answers
Dec 2, 2017

Answer:

#f'(x)=-(3x^2sec^2(2-x^3))/(2sqrt(tan(2-x^3))#

Explanation:

Let #v=-x^3# and #u=tan(2+v)# and #y=f(x)#

Then #y=sqrtu#.

So #f'(x)=dy/dx=dy/(du)*(du)/(dv)*(dv)/(dx)#

#(dv)/dx=-3x^2#

#(du)/(dv)=sec^2(2+v)=sec^2(2-x^3)#

#dy/(du)=1/(2sqrtu)=1/(2sqrt(tan(2-x^3))#

#therefore f'(x)=dy/dx=-3x^2*sec^2(2-x^3)*1/(2sqrt(tan(2-x^3)))#
#=-(3x^2sec^2(2-x^3))/(2sqrt(tan(2-x^3))#

Dec 2, 2017

Answer:

#f'(x)=(-3x^2sec^2(2-x^3))/(2sqrt(tan(2-x^3))#

Explanation:

We will use the chain rule where we take the derivative of the outside and multiply it by the derivative of the inside.

We have #f(x)=sqrt(tan(2-x^3)# It's better to rewrite it as an exponent #(tan(2-x^3))^(1/2)#

So, first lets take the derivative of the #color(blue)("outside")# using the power rule:

#color(blue)(1/2)tan(2-x^3)^color(blue)(1/2-2/2)=> 1/2tan(2-x^3)^(-1/2)#

This is just one part of the answer.

Now let's take the derivative of the #color(green)("inside")#

#1/2(color(green)tan(2-x^3))^(-1/2)#

#d/dxcolor(green)tan(2-x^3) => sec^2(2-x^3)(-3x^2)#

We ended up using the chain rule on the #color(green)("inside")# too.

The derivative of the #tanx# is #sec^2x# but we also have to apply the chain rule here too because #sec^2x# has a function in the inside too. Take the derivative of the #color(red)("function")# inside #sec^2color(red)((2-x^3))# which is where we got the #-3x^2# from.

Now we just multiply both the derivatives of the outside and inside.

#1/(2sqrt(tan(2-x^3)))xxsec^2(2-x^3)(-3x^2)#

#d/dxsqrt(tan(2-x^3)# #=# #(-3x^2sec^2(2-x^3))/(2sqrt(tan(2-x^3))#