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

2 Answers
Dec 2, 2017

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

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))