Find the derivative of
a) y=cos (x/2)
b) yx-3x^2 +6 = y^4 + x
a) -sin (x/2)(2+x)/2^2
-sin(x/2)(x/2)
b) x dy/dx + y - 6x = 4y^3 dy/dx + 3x^2
dy/dx (x -4 y^3) = 3x^2 -y +6x
dy/dx = (3x^2 -y +6x)/(x -4 y^3)
I used inplicit differation for part b
are these correct? Thanks for your help in advance!
your first answer makes no sense.
you are simply finding the derivative of
y = cos (1/2)x
which is
y'=-1/2 sin (x/2)
your second equation:
If your last term of the equation was a typo and it was x^3, then you are correct.
I like the way you showed the first line of your implicit derivative, but according to that line the last term must have been x^3 instead of x.
Let's break down the solutions and verify them:
a) To find the derivative of y = cos(x/2), you correctly applied the chain rule, which states that if you have a composition of functions, you need to multiply the derivative of the outer function by the derivative of the inner function.
So, for y = cos(x/2), the outer function is cosine and the inner function is x/2.
To differentiate cosine, we get -sin(x/2). To differentiate x/2, we get 1/2.
Now, applying the chain rule, we multiply these derivatives:
dy/dx = (-sin(x/2))(1/2) = -sin(x/2)/2.
Therefore, the correct derivative is dy/dx = -sin(x/2)/2. This aligns with your first line and is different from what you provided as the correct answer. It seems that there was a calculation error in your answer.
b) For this problem, you correctly used implicit differentiation. You differentiated each term of the equation with respect to x and then solved for dy/dx.
However, there is a small typo in your equation. Instead of yx - 3x^2 + 6 = y^4 + x, it should be yx - 3x^2 + 6 = y^4 + x^3 (since the power of x should be 3 in the last term).
With this correction, let's go through the steps:
Differentiating yx - 3x^2 + 6 = y^4 + x^3 with respect to x:
d/dx(yx) - d/dx(3x^2) + d/dx(6) = d/dx(y^4) + d/dx(x^3)
Using the product rule, differentiate yx as y(dx/dx) + x(dy/dx), which simplifies to y + x(dy/dx). Also, the derivative of a constant term (such as 6) is 0. And the derivative of a variable term (such as x) is 1.
Therefore, the equation becomes:
y + x(dy/dx) - 6x + 0 = 4y^3(dy/dx) + 3x^2.
Now, solve for dy/dx by isolating the terms with dy/dx:
x(dy/dx) - 4y^3(dy/dx) = 3x^2 - y + 6x.
Factor out dy/dx from the left side:
dy/dx(x - 4y^3) = 3x^2 - y + 6x.
Finally, divide both sides by (x - 4y^3) to isolate dy/dx:
dy/dx = (3x^2 - y + 6x)/(x - 4y^3).
Therefore, the correct expression for the derivative is dy/dx = (3x^2 - y + 6x)/(x - 4y^3). This matches your answer, assuming the last term in the equation was x^3.