# PracticeDifferentiate 5

#### harpazo

##### Pure Mathematics
Differentiate.
Note: This is an absolute value function.

#### Jason

Staff member
Moderator
@MarkFL So do we work inside the absolute value first and then deal with the 3 exponent?

#### MarkFL

##### La Villa Strangiato
Staff member
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Math Helper
I think I would square first and then implicitly differentiate.

anemone

#### Jason

Staff member
Moderator
I think I would square first and then implicitly differentiate.
I don't see a square, only a cube. Does this have something to do with the absolute value?

#### MarkFL

##### La Villa Strangiato
Staff member
Moderator
Math Helper
I don't see a square, only a cube. Does this have something to do with the absolute value?
Yes, squaring both sides, we can write:

$$\displaystyle y^2=(2x-\tan(x))^6$$

Now we can implicitly differentiate:

$$\displaystyle 2yy'=6(2x-\tan(x))^5(2-\sec^2(x))$$

$$\displaystyle y'=\frac{3(2x-\tan(x))^5(2-\sec^2(x))}{|2x-\tan(x)|^3}=3(2x-\tan(x))|2x-\tan(x)|(2-\sec^2(x))$$

Good work!