with respect to x
f(cos x) if f'(x)= 1/x
find (f(cos x))'
i got tan x sec x
this is wrong.
can you explain thank you
To find the derivative of \( f(\cos x) \), we can use the chain rule. The chain rule states that if \( g(x) \) is a function and \( f(x) \) is a differentiable function of \( u \), then \( h(x) = f(g(x)) \) is differentiable and its derivative is given by:
\[ h'(x) = f'(g(x)) \cdot g'(x) \]
In this case, let \( g(x) = \cos x \) and \( f(u) = \dfrac{1}{u} \). To find \((f(\cos x))'\), we need to find \( f'(\cos x) \) and \( (\cos x)' \), and then apply the chain rule.
Step 1: Find \( f'(\cos x) \):
We are given that \( f'(x) = \dfrac{1}{x} \). To find \( f'(\cos x) \), we substitute \( x \) with \( \cos x \):
\[ f'(\cos x) = \dfrac{1}{\cos x} \]
Step 2: Find \( (\cos x)' \):
The derivative of \( \cos x \) with respect to \( x \) is:
\[ (\cos x)' = -\sin x \]
Step 3: Apply the chain rule:
Now we can apply the chain rule by multiplying \( f'(\cos x) \) and \( (\cos x)' \):
\[ (f(\cos x))' = f'(\cos x) \cdot (\cos x)' \]
\[ (f(\cos x))' = \dfrac{1}{\cos x} \cdot (-\sin x) \]
\[ (f(\cos x))' = -\dfrac{\sin x}{\cos x} \]
Therefore, the correct derivative of \( f(\cos x) \) is \( -\dfrac{\sin x}{\cos x} \), which can also be written as \( -\tan x \).