Determine the derivative of the function (² = square):
a) y = (x² + x - 1)Sin²x
b) y = (x² + 1)/Sin²x
chain rule:
d/dx sin^2 x = 2(sinx)(cosx) = sin2x
a)
y = (x^2+x-1)sin^2 x
y' = (2x+1)sin^2 x + (x^2+x-1)sin2x
b)
y = (x^2+1)/sin^2 x
y = (2x*sin^2 x - (x^2+1)*2*sinx*cosx]/sin^4 x
= [2x*sinx - (x^2+1)*cosx]/sin^3 x
Thank you Steve
To find the derivative of these functions, we can use the chain rule and the product rule. The chain rule allows us to differentiate the composite functions while the product rule allows us to differentiate the product of two functions. Let's go through each function step by step.
a) To find the derivative of y = (x² + x - 1)Sin²x:
Step 1: Rewrite the function using the product rule. Let f(x) = (x² + x - 1) and g(x) = Sin²x. Then y(x) = f(x) * g(x).
Step 2: Differentiate f(x) and g(x) separately.
For f(x):
f'(x) = 2x + 1, since the derivative of x² is 2x and the derivative of x is 1.
For g(x):
g'(x) = 2Sin(x) * Cos(x), using the chain rule on Sin²x.
Step 3: Apply the product rule.
Using the product rule, the derivative of y(x) is given by:
y'(x) = f'(x) * g(x) + f(x) * g'(x).
Substituting in the values we found in Step 2:
y'(x) = (2x + 1) * Sin²x + (x² + x - 1) * (2Sin(x) * Cos(x)).
b) To find the derivative of y = (x² + 1)/Sin²x:
Step 1: Rewrite the function using the quotient rule. Let f(x) = x² + 1 and g(x) = Sin²x. Then y(x) = f(x) / g(x).
Step 2: Differentiate f(x) and g(x) separately.
For f(x):
f'(x) = 2x, since the derivative of x² is 2x.
For g(x):
g'(x) = 2Sin(x) * Cos(x), using the chain rule on Sin²x.
Step 3: Apply the quotient rule.
Using the quotient rule, the derivative of y(x) is given by:
y'(x) = (f'(x) * g(x) - f(x) * g'(x)) / (g(x))².
Substituting in the values we found in Step 2:
y'(x) = (2x * Sin²x - (x² + 1) * (2Sin(x) * Cos(x))) / (Sin²x)².
These are the derivatives of the given functions.