Differentiate implicitly (a) x^3+y^3=27xy
x^3+y^3=27xy
I assume you are differentiating with respect to x.
3x^2 + 3y^2 dy/dx = 27x dy/dx + 27y
3y^2 dy/dx - 27x dy/dx = 27y - 3x^2
dy/dx (3y^2 - 27x) = 27y - 3x^2
dy/dx = (27y - 3x^2)/(3y^2 - 27x)
= (9y - x^2)/(y^2 - 9x) or (x^2 - 9y)/(9x - y^2)
check my algebra, did not write it out first
Sure, let me differentiate that for you implicitly, but beware, it may get a bit derivative!
Using the chain rule, we differentiate each term with respect to x:
For the x^3 term, we get 3x^2.
For the y^3 term, we get 3y^2 * dy/dx.
For the 27xy term, we get 27y + 27x * dy/dx.
Now let's equate these derivatives to zero, because that's where the fun begins!
3x^2 + 3y^2 * dy/dx = 27y + 27x * dy/dx.
To find dy/dx, let's solve this equation step-by-step, with a side of laughter:
First, let's gather all the dy/dx terms on one side, to form a delightful rollercoaster of derivatives:
3y^2 * dy/dx - 27x * dy/dx = 27y - 3x^2.
Now, let's factor out the dy/dx:
dy/dx * (3y^2 - 27x) = 27y - 3x^2.
To make it easier on the eyes (and funnier), let's simplify the equation further:
dy/dx = (27y - 3x^2) / (3y^2 - 27x).
And there you have it, the hilariously derived derivative dy/dx for the equation x^3 + y^3 = 27xy.
Now that we've gone through all that math humor, don't forget, you can always count on me to bring the chuckles!
To differentiate implicitly, we will differentiate both sides of the equation with respect to x. When differentiating, we treat y as a function of x and use the chain rule.
Let's begin step by step.
Step 1: Begin by differentiating both sides of the equation with respect to x.
d/dx(x^3 + y^3) = d/dx(27xy)
Step 2: Differentiate each term separately.
d/dx(x^3) + d/dx(y^3) = d/dx(27xy)
Step 3: Apply the power rule to differentiate x^3.
3x^2 + d/dx(y^3) = 27(d/dx(xy))
Step 4: Differentiate y^3 using the chain rule.
3x^2 + 3y^2 * (dy/dx) = 27 * (xdy/dx + y)
Step 5: Rearrange the equation to isolate (dy/dx) term.
3x^2 - 27xdy/dx = 27y - 3y^2(dy/dx)
Step 6: Simplify the equation by grouping similar terms.
(3x^2 + 3y^2) * (dy/dx) = 27y - 27xy
Step 7: Divide both sides by (3x^2 + 3y^2).
(dy/dx) = (27y - 27xy) / (3x^2 + 3y^2)
That's the implicit differentiation of the equation x^3 + y^3 = 27xy.
To differentiate a function implicitly, we differentiate both sides of the equation with respect to the variable you want to differentiate with respect to. In this case, we want to differentiate with respect to both x and y.
Let's differentiate both sides of the equation x^3 + y^3 = 27xy with respect to x:
Taking the derivative of x^3 with respect to x, we use the power rule:
d/dx (x^3) = 3x^2
Taking the derivative of y^3 with respect to x is a bit tricky since y is considered as a function of x. Therefore, we use the chain rule:
d/dx (y^3) = d/dy (y^3) * dy/dx
To find dy/dx, we can differentiate the original equation implicitly with respect to y instead of x:
Differentiating both sides of the equation x^3 + y^3 = 27xy with respect to y:
Taking the derivative of x^3 with respect to y is zero since x is considered a constant:
d/dy (x^3) = 0
Taking the derivative of y^3 with respect to y using the power rule:
d/dy (y^3) = 3y^2
Now, let's solve for dy/dx by rearranging the equation from the previous step:
3y^2 * dy/dx = 27x - 3x^2
dy/dx = (27x - 3x^2) / (3y^2)
Now, substitute this expression for dy/dx back into the equation we started with to get the complete derivative:
3x^2 + (27x - 3x^2) / (3y^2) = 27
Simplifying this equation will give you the derivative implicitly in terms of x and y.