find the 2nd derivative of
y= 1/x +cosxsinx
I got 2/x^3 - 2cosxsinx
is this right or did I get this wrong
To find the second derivative of the function y = 1/x + cos(x)sin(x), we will need to differentiate it twice.
First, let's find the first derivative using the sum rule and the product rule:
For the term 1/x:
- Using the power rule, the derivative of 1/x is -1/x^2.
For the term cos(x)sin(x):
- Using the product rule, the derivative is [-sin(x)]sin(x) + cos(x)[cos(x)].
- Simplifying this, we get -sin^2(x) + cos^2(x).
Now, combining the derivatives, we have:
dy/dx = -1/x^2 + (-sin^2(x) + cos^2(x))
To find the second derivative, we take the derivative of this expression again:
For the term -1/x^2:
- Applying the power rule, we find that the derivative is 2/x^3.
For the term -sin^2(x) + cos^2(x):
- Using the chain rule, the derivative of sin^2(x) is 2sin(x)cos(x).
- Similarly, the derivative of cos^2(x) is -2cos(x)sin(x).
- Combining these, we get 2sin(x)cos(x) - 2cos(x)sin(x) = 0.
Therefore, the second derivative (d^2y/dx^2) is equal to 2/x^3 + 0, which simplifies to 2/x^3.
So, your answer of 2/x^3 is correct. Well done!