Practice Differentiate 5

harpazo

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Differentiate.
Note: This is an absolute value function.

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TheJason

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

MarkFL

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I think I would square first and then implicitly differentiate.
 
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TheJason

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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

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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:

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

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

harpazo

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Good work!