Determine the first second and third derivatives of sin2x^2
y = sin(2x^2)
y' = 4x cos(2x^2)
y" = 4cos(2x^2) - 16x^2 sin(2x^2)
y"' = -48x sin(2x^2) - 64x^3 cos(2x^2)
To find the derivatives of sin(2x^2), we can start by using the chain rule.
First derivative:
y = sin(2x^2)
y' = cos(2x^2) * d/dx (2x^2)
= cos(2x^2) * 4x
= 4x cos(2x^2)
Second derivative:
y' = 4x cos(2x^2)
y" = (4 cos(2x^2)) * d/dx (4x) + (4x) * d/dx (cos(2x^2))
= 4 cos(2x^2) * 4 + 4x * (-2sin(2x^2)) * d/dx (2x^2)
= 16 cos(2x^2) - 8x sin(2x^2)
Third derivative:
y" = 16 cos(2x^2) - 8x sin(2x^2)
y"' = (16 cos(2x^2) - 8x sin(2x^2)) * d/dx (2x^2) + (-8sin(2x^2)) * d/dx (2x^2) + (-8x) * d/dx (sin(2x^2))
= (16 cos(2x^2) - 8x sin(2x^2)) * 4x + (-8sin(2x^2)) * 4x + (-8x) * (2cos(2x^2)) * d/dx (2x^2)
= 64x^2 cos(2x^2) - 32x^2 sin(2x^2) - 16x cos(2x^2)
= -16x cos(2x^2) - 32x^2 sin(2x^2) + 64x^2 cos(2x^2)
= -16x cos(2x^2) - 32x^2 sin(2x^2) + 64x^2 cos(2x^2)
= -48x sin(2x^2) - 64x^3 cos(2x^2)
Therefore, the first derivative is y' = 4x cos(2x^2),
the second derivative is y" = 16 cos(2x^2) - 8x sin(2x^2),
and the third derivative is y"' = -48x sin(2x^2) - 64x^3 cos(2x^2).
To find the derivatives of the given function, sin^2(x^2), we will use the chain rule and derivatives of trigonometric functions. Let's begin:
Step 1: First Derivative
To find the first derivative, we apply the chain rule to the function sin^2(x^2).
Let u = x^2
Then y = sin^2(u)
Now, we can find dy/du by taking the derivative of y with respect to u:
dy/du = 2sin(u)cos(u) = 2sin(u)cos(u) = 2sin(x^2)cos(x^2)
Finally, we can find dy/dx by multiplying dy/du by du/dx:
dy/dx = (dy/du)(du/dx) = 2sin(x^2)cos(x^2)(2x) = 4xsin(x^2)cos(x^2)
Therefore, the first derivative of sin^2(x^2) is 4xsin(x^2)cos(x^2).
Step 2: Second Derivative
To find the second derivative, we differentiate the first derivative with respect to x.
d^2y/dx^2 = d/dx(4xsin(x^2)cos(x^2))
Using the product rule, we have:
d^2y/dx^2 = 4cos(x^2)cos(x^2) + 4xsin(x^2)(-2sin(x^2)x)
Simplifying further:
d^2y/dx^2 = 4cos^2(x^2) - 8x^2sin^2(x^2)
Therefore, the second derivative of sin^2(x^2) is 4cos^2(x^2) - 8x^2sin^2(x^2).
Step 3: Third Derivative
To find the third derivative, we differentiate the second derivative with respect to x.
d^3y/dx^3 = d/dx(4cos^2(x^2) - 8x^2sin^2(x^2))
Differentiating each term separately:
d^3y/dx^3 = 0 - 8sin(2x^2)(2x) - 8sin^2(x^2)(4x) + 16xsin(x^2)(2sin(x^2)x)
Simplifying further:
d^3y/dx^3 = -16x^2sin(2x^2) - 32x^2sin^2(x^2) + 32x^2sin(x^2)sin^3(x^2)
Therefore, the third derivative of sin^2(x^2) is -16x^2sin(2x^2) - 32x^2sin^2(x^2) + 32x^2sin(x^2)sin^3(x^2).
And that's the first, second, and third derivatives of sin^2(x^2).
To determine the derivatives of the function \( \sin^2(x^2) \), we can follow these steps:
Step 1: Find the first derivative.
Step 2: Find the second derivative.
Step 3: Find the third derivative.
Let's go through each step:
Step 1: Find the first derivative.
To find the first derivative, we can apply the chain rule. The chain rule states that if we have a function \( f(g(x)) \), then its derivative is given by \( f'(g(x)) \cdot g'(x) \).
In this case, we have \( f(x) = \sin^2(x^2) \), so we need to find the derivative of \( f(x) \). Let's break it down:
Let \( g(x) = x^2 \).
\( f(x) = \sin^2(g(x)) \)
Now, let's find the derivative of \( f(x) \) step by step:
\( f'(x) = 2 \sin(g(x)) \cdot \cos(g(x)) \cdot g'(x) \)
Since \( g(x) = x^2 \), we find:
\( g'(x) = 2x \)
Finally, substituting everything back into \( f'(x) \):
\( f'(x) = 2 \sin(x^2) \cdot \cos(x^2) \cdot 2x \)
So, the first derivative is:
\( \frac{d}{dx}(\sin^2(x^2)) = 4x \sin(x^2) \cdot \cos(x^2) \)
Step 2: Find the second derivative.
To find the second derivative, we will differentiate the first derivative \( f'(x) \) with respect to \( x \). Applying the product rule and chain rule, we get:
\( f''(x) = \frac{d}{dx}(4x \cdot \sin(x^2) \cdot \cos(x^2)) \)
Simplifying this expression, we find:
\( f''(x) = 4 \cos^2(x^2) - 4x^2 \sin^2(x^2) \)
So, the second derivative is:
\( \frac{d^2}{dx^2}(\sin^2(x^2)) = 4 \cos^2(x^2) - 4x^2 \sin^2(x^2) \)
Step 3: Find the third derivative.
To find the third derivative, we will differentiate the second derivative \( f''(x) \) with respect to \( x \) again. Applying the product rule and chain rule, we get:
\( f'''(x) = \frac{d}{dx}(4 \cos^2(x^2) - 4x^2 \sin^2(x^2)) \)
Simplifying this expression, we find:
\( f'''(x) = -8x \sin(x^2) \cos(x^2) - 16x \sin^3(x^2) + 8x \cos^3(x^2) \)
So, the third derivative is:
\( \frac{d^3}{dx^3}(\sin^2(x^2)) = -8x \sin(x^2) \cos(x^2) - 16x \sin^3(x^2) + 8x \cos^3(x^2) \)
And there you have it! You now know the first, second, and third derivatives of \( \sin^2(x^2) \).