Differentiate: a) y= x^2/ sinx
b) y= In(2x^3+x)
I will be happy to critique your work here. These are standard formulas.
a) =2x^1(x^2)/xcos(sinx)?
I can't see you you came up with that very strange looking answer.
As bobpursley pointed out, this is a direct application of the quotient rule
I got
dy/dx = (2x sinx - x^2 cosx)/(sin^2 x)
simplification would leave to several final versions of an answer, certainly not yours.
the 2nd is even easier.
dy/dx = (6x^2+1)/(2x^3 + x)
To differentiate the given functions, we'll use the rules of differentiation. Here's how you can differentiate each function:
a) y = (x^2)/sin(x)
To differentiate this function, we can use the quotient rule which states:
If we have a function p(x) divided by a function q(x), where p(x) and q(x) are differentiable functions, then the derivative of p(x)/q(x) is given by:
(p'(x)q(x) - p(x)q'(x))/(q(x))^2
In our case, p(x) = x^2 and q(x) = sin(x). Let's find the derivatives of p(x) and q(x) first:
p'(x) = 2x (using the power rule, since x^2 is a polynomial)
q'(x) = cos(x) (using the derivative of sin(x) which is cos(x))
Now, substituting these values into the quotient rule formula:
y' = (2x*sin(x) - x^2*cos(x))/(sin(x))^2
So, the derivative of y = (x^2)/sin(x) is (2x*sin(x) - x^2*cos(x))/(sin(x))^2.
b) y = ln(2x^3 + x)
To differentiate the natural logarithm function, we use the chain rule. The chain rule states that if we have a function f(g(x)), then the derivative of f(g(x)) is given by:
(f'(g(x)))*(g'(x))
In our case, f(x) = ln(x) and g(x) = 2x^3 + x. Let's find the derivatives of f(x) and g(x) first:
f'(x) = 1/x (using the derivative of ln(x) which is 1/x)
g'(x) = 6x^2 + 1 (by applying the power rule to 2x^3 and the derivative of x, which is 1)
Now, substituting these values into the chain rule formula:
y' = (1/(2x^3 + x)) * (6x^2 + 1)
So, the derivative of y = ln(2x^3 + x) is (6x^2 + 1)/(2x^3 + x).