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How do you differentiate #ln(xy)=cos(y^4)#?

Answer 1

Paisley Johnson

#(dy)/(dx)=-y/(x(4y^4sin(y^4)+1)#

#ln(xy)=cos(y^4)#

#lnx+lny=cos(y^4)#

differentiate wrt #x#

#d/(dx)(lnx+lny)=d/(dx)(cosy^4)#

#1/x+1/y(dy)/(dx)=4y^3(dy)/(dx)(-sin(y^4))#

#4y^3(dy)/(dx)sin(y^4)+1/y(dy)/(dx)=-1/x#

#(dy)/(dx)(4y^3sin(y^4)+1/y)=-1/x#

#(dy)/(dx)((4y^4sin(y^4)+1)/y)=-1/x#

#(dy)/(dx)=-y/(x(4y^4sin(y^4)+1)#

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Answer 2

Axel Alvarez

To differentiate ln(xy) = cos(y^4), you would use the chain rule. First, differentiate ln(xy) with respect to both x and y separately, then differentiate cos(y^4) with respect to y, and finally apply the chain rule to the product ln(xy).

The differentiation of ln(xy) with respect to x yields (1/y), and with respect to y yields (ln(xy))'. For cos(y^4), the derivative with respect to y is -4y^3*sin(y^4).

So, using the chain rule, the differentiation of ln(xy) is (1/y) * (dx/dy) + (ln(xy)) * (dy/dy), and for cos(y^4), it is -4y^3*sin(y^4) * (dy/dx).

Then, you can solve for dx/dy and dy/dx to find the partial derivatives with respect to x and y, respectively.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.