Need help to find dy/dx:
1) y = (sin x) ^-1/3
answer dy/dx = - 1/ 3(sinx) ^4/3
2) y = sqrt 8 + sin(6x)
answer - lost
1) close, you left out the derivative of the base, so
dy/dx = - 1/ 3(sinx) ^4/3(cosx)
for 2)
rewrite it as
y = (8+sin(6x))^1/2
dy/dx = (1/2)((8+sin(6x))^-1/2)(cos(6x))(6)
= 3cos(6x)((8+sin(6x))^-1/2)
To find the derivative of the given function, we can use the rules of differentiation. Let's go through each problem step by step:
1) To find dy/dx of y = (sinx)^(-1/3), we can use the chain rule.
First, let's define our function as u = sinx.
Now, we can rewrite y as y = u^(-1/3).
To find dy/du, we can use the power rule, which states that if y = x^n, then dy/dx = n * x^(n-1):
dy/du = (-1/3) * u^(-1/3 - 1) = (-1/3) * u^(-4/3).
Next, we need to multiply dy/du by du/dx (the derivative of u with respect to x) to find dy/dx:
du/dx = cosx (since the derivative of sinx with respect to x is cosx).
Now, we can find dy/dx:
dy/dx = du/dx * dy/du
= cosx * (-1/3) * (sinx)^(-4/3)
= - 1/ 3(sinx) ^4/3
Therefore, dy/dx = - 1/ 3(sinx) ^4/3.
2) To find dy/dx of y = sqrt(8) + sin(6x), we can differentiate each term separately.
The derivative of the constant term sqrt(8) is zero since the derivative of a constant is always zero.
To find the derivative of the second term, sin(6x), we can use the chain rule. Let u = 6x, so we have y = sin(u).
Using the chain rule, dy/du = cos(u) * du/dx = cos(6x) * 6.
Therefore, dy/dx = 0 + cos(6x) * 6 = 6cos(6x).
Hence, the derivative of y = sqrt(8) + sin(6x) is dy/dx = 6cos(6x).