find the first, second and third derivative of : sin3x^2?
just use the chain rule
y = sin(3x^2)
y' = cos(3x^2)(6x)
y" = -sin(3x^2)(6x)^2 + cos(3x^2)(6)
y"' = -cos(3x^2)(6x)^3 - sin(3x^2)(2*6x)(6) - sin(3x^2)(6x)(6)
I'll let you do the algebra. You can always check your results by using wolframalpha.com
Just type 3rd derivative sin(3x^2)
Just look for the first,second and third derivative
To find the first derivative of the function \(f(x) = \sin(3x^2)\), we can use the chain rule. Here's the step-by-step process:
Step 1: Differentiate the outer function with respect to the inner function multiplied by the derivative of the inner function.
\(f'(x) = \cos(3x^2) \cdot \frac{d}{dx}(3x^2)\)
Step 2: Differentiate the inner function \(3x^2\).
\(f'(x) = \cos(3x^2) \cdot 6x\)
So, the first derivative of \(\sin(3x^2)\) is \(f'(x) = \cos(3x^2) \cdot 6x\).
Now, let's find the second derivative by applying the product rule:
Step 3: Differentiate the first term \(\cos(3x^2)\) with respect to \(x\) and leave the second term \(6x\) as is. Then add the second term \(6x\) differentiated with respect to \(x\) multiplied by the first term \(\cos(3x^2\).
\(f''(x) = \frac{d}{dx}(\cos(3x^2)) \cdot 6x + \cos(3x^2) \cdot \frac{d}{dx}(6x)\)
Step 4: Determine the derivative of the first term \(\cos(3x^2)\) and the derivative of the second term \(6x\).
\(f''(x) = -2x \sin(3x^2) + 6 \cos(3x^2)\)
Thus, the second derivative of \(\sin(3x^2)\) is \(f''(x) = -2x \sin(3x^2) + 6 \cos(3x^2)\).
Lastly, let's find the third derivative by applying the product rule again:
Step 5: Differentiate the first term \(-2x \sin(3x^2)\) with respect to \(x\) and leave the second term \(6 \cos(3x^2)\) as is. Then add the second term \(6 \cos(3x^2)\) differentiated with respect to \(x\) multiplied by the first term \(-2x \sin(3x^2)\).
\(f'''(x) = \frac{d}{dx}(-2x \sin(3x^2)) \cdot 6 \cos(3x^2) + (-2x \sin(3x^2)) \cdot \frac{d}{dx}(6 \cos(3x^2))\)
Step 6: Determine the derivative of the first term \(-2x \sin(3x^2)\) and the derivative of the second term \(6 \cos(3x^2)\).
\(f'''(x) = (-2 \sin(3x^2) - 6x \cos(3x^2)) \cdot 6 \cos(3x^2) + (-2x \sin(3x^2)) \cdot (-18x \sin(3x^2))\)
Simplifying the expression yields:
\(f'''(x) = -12x \sin^2(3x^2) - 72x^2 \sin(3x^2) \cos(3x^2)\)
Thus, the third derivative of \(\sin(3x^2)\) is \(f'''(x) = -12x \sin^2(3x^2) - 72x^2 \sin(3x^2) \cos(3x^2)\).
To find the first, second, and third derivative of sin(3x^2), we will follow these steps:
Step 1: Find the derivative of sin(3x^2) with respect to x.
Step 2: Differentiate the result from step 1 to find the second derivative.
Step 3: Finally, differentiate the second derivative to find the third derivative.
Let's begin with step 1:
Step 1: Find the first derivative
To find the derivative of sin(3x^2) with respect to x, we will use the chain rule, which states that if we have a function of the form f(g(x)), the derivative is equal to f'(g(x)) * g'(x).
Here, f(g(x)) = sin(g(x)) and g(x) = 3x^2.
To find f'(g(x)), we differentiate sin(x) with respect to x, which is cos(x).
To find g'(x), we differentiate 3x^2 with respect to x, which is 6x.
Now, let's apply the chain rule:
The first derivative of sin(3x^2) will be:
f'(g(x)) * g'(x) = cos(g(x)) * g'(x) = cos(3x^2) * 6x.
So, the first derivative (dy/dx) of sin(3x^2) is:
dy/dx = cos(3x^2) * 6x.
Moving on to step 2:
Step 2: Find the second derivative
To find the second derivative, we need to differentiate the first derivative with respect to x.
Differentiating dy/dx = cos(3x^2) * 6x with respect to x:
Applying the product rule, the derivative of cos(3x^2) * 6x will be:
(cos(3x^2))' * 6x + cos(3x^2) * (6x)'.
Differentiating cos(3x^2) gives us: -sin(3x^2) * (3x^2)' = -sin(3x^2) * 6x.
Differentiating 6x gives us: 6.
So, the second derivative (d^2y/dx^2) of sin(3x^2) is:
d^2y/dx^2 = -sin(3x^2) * 6x + 6.
Now, let's move on to step 3:
Step 3: Find the third derivative
To find the third derivative, we need to differentiate d^2y/dx^2 = -sin(3x^2) * 6x + 6 with respect to x.
Differentiating -sin(3x^2) * 6x gives us: (-sin(3x^2))' * 6x + -sin(3x^2) * (6x)'.
The derivative of -sin(3x^2) is: -cos(3x^2) * (3x^2)' = -cos(3x^2) * 6x.
Differentiating 6 gives us: 0.
So, the third derivative (d^3y/dx^3) of sin(3x^2) is:
d^3y/dx^3 = -cos(3x^2) * 6x + 0.
Simplifying, we obtain:
d^3y/dx^3 = -6x*cos(3x^2).
Thus, the first, second, and third derivatives of sin(3x^2) are:
First derivative (dy/dx) = cos(3x^2) * 6x.
Second derivative (d^2y/dx^2) = -sin(3x^2) * 6x + 6.
Third derivative (d^3y/dx^3) = -6x * cos(3x^2).