Hello everyone:
I am wondering if it is possible to use both the chain rule and product rule to differentiate something.
For example, if I am trying to find the second derivative of \(\displaystyle f(x) = \cos x\):
\(\displaystyle f(x) = \cos x\)
\(\displaystyle f'(x) = -\sin x\)
Applying the Product Rule:
\(\displaystyle f''(x) = 0 + (-1)(\cos x)\)
\(\displaystyle f''(x) = -\cos x\)
When I apply the Chain Rule:
\(\displaystyle f(x) = \sin x\), \(\displaystyle f'(x) = cos x\)
\(\displaystyle g(x) = -x\), \(\displaystyle g'(x) = -1\)
Therefore:
\(\displaystyle f''(x) = -\cos(-x)\)
This is different from when I got by using the product rule.
I am wondering if it is possible to use both the chain rule and product rule to differentiate something.
For example, if I am trying to find the second derivative of \(\displaystyle f(x) = \cos x\):
\(\displaystyle f(x) = \cos x\)
\(\displaystyle f'(x) = -\sin x\)
Applying the Product Rule:
\(\displaystyle f''(x) = 0 + (-1)(\cos x)\)
\(\displaystyle f''(x) = -\cos x\)
When I apply the Chain Rule:
\(\displaystyle f(x) = \sin x\), \(\displaystyle f'(x) = cos x\)
\(\displaystyle g(x) = -x\), \(\displaystyle g'(x) = -1\)
Therefore:
\(\displaystyle f''(x) = -\cos(-x)\)
This is different from when I got by using the product rule.