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Mathematics 15 Online
OpenStudy (anonymous):

If 6x^2+xy-cos(y)=10, what is dy/dx?

OpenStudy (mathmale):

EJ: As you'll see, this equation has NOT been solved for y, and it would be quite difficult to solve it for y. therefore, to find the derivative dy/dx, use IMPLICIT DIFFERENTIATION.

OpenStudy (mathmale):

Enclose your original equation in parentheses and apply the derivative operator (d/dx) to the whole thing. Differentiate term by term. Note that xy is a product, which means you MUST apply the product rule, while remembering that y is presumed to be a function of x.

OpenStudy (mathmale):

EJ?

OpenStudy (anonymous):

Thank you! Can you explain getting the derivative of the function? There are a lot of terms.

OpenStudy (mathmale):

I can guide you thru the process. First, what is the derivative (with respect to x) of 6x^2?

OpenStudy (anonymous):

12x

OpenStudy (mathmale):

Right. Given that xy is a product, and that y is assumed to be a function of x, what is the derivative of xy? ( )( )+( )( )

OpenStudy (anonymous):

(x)(dy/dx)+(y)(dx/dx)

OpenStudy (mathmale):

Yes, very good. And dx/dx=?

OpenStudy (anonymous):

what about -cos(y)?

OpenStudy (anonymous):

dx/dx=1

OpenStudy (mathmale):

First: dx/dx= 1. so your derivative of xy is therefore x(dy/dx)+y. OK?

OpenStudy (anonymous):

yes

OpenStudy (mathmale):

What is the derivative of -cos y? Recall that y is assumed to be a function of x, so we must use the Chain Rule and differentiate y. What do you recall about the derivative of cos x? (d/dx)(cos x) = ?

OpenStudy (anonymous):

-sinx

OpenStudy (mathmale):

Right, and so the derivative of -cos x is what?

OpenStudy (anonymous):

sinx

OpenStudy (mathmale):

Right. Now, knowing that y is supposedly a function of x, what is the derivative with respect to x of -cos y?

OpenStudy (anonymous):

sinx (dy/dx)

OpenStudy (mathmale):

That's great! right on target!

OpenStudy (mathmale):

What's the derivative of 10?

OpenStudy (anonymous):

perfect thank you! Then 10 goes to zero and i solve for dy/dx right?

OpenStudy (mathmale):

That's right. Confident that you can finish up?

OpenStudy (anonymous):

yes thank you so much

OpenStudy (mathmale):

My great pleasure, EJ! "See" you again on Open Study.

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