What is the derivative of:
y = sin(1)sin^2(x^4)+sec^2(pix^2+x+1)
sin.45
sinx
sinx=1
To find the derivative of the given expression, we can use the chain rule and the product rule. Let's break it down step by step:
First, we have two terms in the expression:
Term 1: sin(1)sin^2(x^4)
Term 2: sec^2(pix^2+x+1)
Let's differentiate each term separately:
For Term 1, y1 = sin(1)sin^2(x^4):
1. Differentiating sin(1) gives us 0, as sin(1) is a constant.
2. To differentiate sin^2(x^4), we apply the chain rule. Let u = sin(x), so we have u^2. The derivative is 2u * du/dx = 2sin(x) * cos(x).
So, the derivative of Term 1 is 0 + 2sin(x) * cos(x) * d(x^4)/dx = 2sin(x) * cos(x) * 4x^3 = 8x^3 * sin(x) * cos(x).
For Term 2, y2 = sec^2(pix^2+x+1):
1. To differentiate sec^2(u), where u = pix^2 + x + 1, we use the chain rule. The derivative is 2 * sec(u) * tan(u) * du/dx.
2. du/dx = d(pix^2 + x + 1)/dx = (2pix + 1)
3. Therefore, the derivative of Term 2 is 2 * sec(pix^2 + x + 1) * tan(pix^2 + x + 1) * (2pix + 1).
To find the derivative of the entire expression, y = Term 1 + Term 2, we sum the derivatives of the two terms:
dy/dx = 8x^3 * sin(x) * cos(x) + 2 * sec(pix^2 + x + 1) * tan(pix^2 + x + 1) * (2pix + 1).
That's the derivative of y = sin(1)sin^2(x^4) + sec^2(pix^2+x+1).