Find second differentiation for function x^3=ycosx
I assume you differentiating with respect to x ?
3x^2 = y(-sinx) + dy/dx(cosx)
dy/dx = 3x^2 + ysinx
d(dy/dx)/dx = 6x + dy/dx sinx + ycosx
= 6x + (3x^2 + ysinx)sinx + ycosx
= 6x^2 + 3x^2sinx + ysin^2 x + ycosx
check my work, should have written it on paper first
or
y = x^3/cosx
y' = (cosx(3x^2) - x^3(-sinx))/cos^2x
= (3x^2 cosx + x^3 sinx)/cos^2x
now do it again
Why 6x become power of 2 at the end?
That was clearly a typo, good for you to catch it
To find the second derivative of the function y = x^3 * cos(x), we need to follow these steps:
Step 1: Find the first derivative.
To find the first derivative, we will use the product rule. The product rule states that if y = u * v, where both u and v are functions of x, then the first derivative of y is given by the formula:
dy/dx = u * dv/dx + v * du/dx
In this case, u = x^3 and v = cos(x). Taking the derivatives, we have:
du/dx = 3x^2 (derivative of x^3)
dv/dx = -sin(x) (derivative of cos(x))
Substituting these values into the product rule formula, we get:
dy/dx = x^3 * (-sin(x)) + cos(x) * 3x^2
Step 2: Simplify the expression.
To simplify the expression, we can distribute and combine like terms:
dy/dx = -x^3 * sin(x) + 3x^2 * cos(x)
Step 3: Find the second derivative.
To find the second derivative, we need to take the derivative of the first derivative. Using the product rule again, we have:
d^2y/dx^2 = (d/dx)(-x^3 * sin(x) + 3x^2 * cos(x))
Taking the derivatives of the individual terms, we get:
(d/dx)(-x^3 * sin(x)) = -3x^2 * sin(x) - x^3 * cos(x)
(d/dx)(3x^2 * cos(x)) = 6x * cos(x) - 3x^2 * sin(x)
Substituting these values back into the expression, we have:
d^2y/dx^2 = -3x^2 * sin(x) - x^3 * cos(x) + 6x * cos(x) - 3x^2 * sin(x)
Simplifying further, we can combine like terms:
d^2y/dx^2 = -4x^2 * sin(x) + 6x * cos(x) - x^3 * cos(x)
So, the second derivative of the function y = x^3 * cos(x) is given by -4x^2 * sin(x) + 6x * cos(x) - x^3 * cos(x).