Differentiate:
y = sin(2x - 1) + cos(x^3) - tan(3x)
Do each term, using your basic trig derivatives
so
y' = 2cos(2x-1) -3x^2(sin(x^3)) - 3sec2 (3x)
Differentiate y=cos(3x)sin(x/2)
To differentiate the given expression, we will use the basic rules of differentiation.
Rule 1: The derivative of a constant is zero.
Rule 2: The derivative of x^n is n*x^(n-1).
Rule 3: The derivative of sin(x) is cos(x).
Rule 4: The derivative of cos(x) is -sin(x).
Rule 5: The derivative of tan(x) is sec^2(x), where sec(x) = 1/cos(x).
Now, let's differentiate each term one by one.
Term 1: y = sin(2x - 1)
To differentiate sin(2x - 1), we need to apply the chain rule. The chain rule states that if we have a composite function like f(g(x)), its derivative is given by f'(g(x)) * g'(x).
Let's denote u = 2x - 1, so y = sin(u).
Using the chain rule, dy/du = cos(u)
Now, we need to differentiate u = 2x - 1 with respect to x, which gives du/dx = 2.
Multiplying the two derivatives together, we get dy/dx = cos(u) * 2.
Substituting back u = 2x - 1, we have dy/dx = cos(2x - 1) * 2.
Term 2: y = cos(x^3)
To differentiate cos(x^3), we need to apply the chain rule again.
Let's denote v = x^3, so y = cos(v).
Using the chain rule, dy/dv = -sin(v)
Now, we need to differentiate v = x^3 with respect to x, which gives dv/dx = 3x^2.
Multiplying the two derivatives together, we get dy/dx = -sin(x^3) * 3x^2.
Term 3: y = tan(3x)
The derivative of tan(3x) is directly given by the rule 5.
dy/dx = sec^2(3x) * 3.
Note: If you explicitly want to write the derivative in terms of cos and sin, you can use the identity sec^2(x) = 1/cos^2(x) and simplify accordingly.
Now, summing up all the derivatives from the three terms, the total derivative dy/dx is:
dy/dx = cos(2x - 1) * 2 + (-sin(x^3) * 3x^2) + sec^2(3x) * 3.
So, the final derivative of y = sin(2x - 1) + cos(x^3) - tan(3x) is dy/dx = cos(2x - 1) * 2 + (-sin(x^3) * 3x^2) + sec^2(3x) * 3.