What is the slope of the tangent line of #tan(xy)-cot(x^2)= C #, where C is an arbitrary constant, at #(pi/3,pi/3)#?

Question

Emilia Aguilar · Accepted Answer

#d/dx[tan(xy)-cot(x^2)=C]#

We will use implicit differentiation and the chain rule next.

#stackrel"product rule"overbrace(d/dx[xy])sec^2(xy)-d/dx[x^2]*-csc^2(x^2)=0#

#sec^2(xy)*(y+x[dy/dx])+2xcsc^2(x^2)=0#

We can now plug in #(pi/3,pi/3)# and solve for #dy/dx#, which will give us the slope of the tangent line at that point.

#sec^2(pi^2/9)*(pi/3+pi/3[dy/dx])+(2pi)/3csc^2((pi^2)/9)=0#

From here, it's just algebra.

William Abdalla · Answer

undefined To find the slope of the tangent line at the point (pi/3, pi/3), we first need to find the derivative of the given function with respect to both x and y. Then, we can evaluate the derivative at the point (pi/3, pi/3) to find the slope of the tangent line.

Differentiating the function tan(xy) - cot(x^2) = C with respect to x using implicit differentiation, we get:

d/dx [tan(xy)] - d/dx [cot(x^2)] = 0

Applying the chain rule and product rule, the derivative of tan(xy) with respect to x is:

sec^2(xy) * (y + x(dy/dx))

And the derivative of cot(x^2) with respect to x is:

-d/dx [1/tan(x^2)] = -(-2x/sin^2(x^2)) = 2x/sin^2(x^2)

Plugging these derivatives back into the equation and solving for dy/dx, we get:

sec^2(xy) * (y + x(dy/dx)) + 2x/sin^2(x^2) = 0

Now, evaluate this expression at the point (pi/3, pi/3) to find the slope of the tangent line at that point.