Derivatives of Trig Functions Vs Those Without

[Jason76]

\(\displaystyle y' = \ln(\cos 2x) = \dfrac{1}{\cos2x} * (-\sin2x) * 2\)

On the other hand...

\(\displaystyle y' = \ln(7x - 14) = \dfrac{1}{7x - 14} * 7\)

Why the difference? It seems that \(\displaystyle \ln(cos2x)\) is differentiated twice and this is multiplied to the original problem. While, in the other one the source problem is differentiated once and this is multiplied to the source problem.

 

[pka]

\(\displaystyle y' = \ln(\cos 2x) = \dfrac{1}{\cos2x} * (-\sin2x) * 2\)

On the other hand...

\(\displaystyle y' = \ln(7x - 14) = \dfrac{1}{7x - 14} * 7\)

Why the difference? It seems that \(\displaystyle \ln(cos2x)\) is differentiated twice and this is multiplied to the original problem. While, in the other one the source problem is differentiated once and this is multiplied to the source problem.

If \(\displaystyle y=\ln(\cos(x))\) then \(\displaystyle y'=\dfrac{1}{\cos(x)}(-\sin(x))\).

Now \(\displaystyle (\cos(2x))'=(-2\sin(2x))\)

Now what is your point?

 

[Jason76]

Well some might think that, considering u * du, the answer would be \(\displaystyle \dfrac{1}{\cos2x} * -\sin2\) or \(\displaystyle \dfrac{1}{\cos2x} * -2\sin\)

 

[pka]

Well some might think that, considering d * du, the answer would be \(\displaystyle \dfrac{1}{\cos2x} * -\sin2\)

Why would anyone think that?

If \(\displaystyle y=\ln(f(g(h(x)))\) then \(\displaystyle y'=\dfrac{1}{f(g(h(x))}(f'(g(h(x))(g'(h(x))(h'(x))\).

Simple chain rule.

 

[DrPhil]

\(\displaystyle y' = \ln(\cos 2x) = \dfrac{1}{\cos2x} * (-\sin2x) * 2\)

On the other hand...

\(\displaystyle y' = \ln(7x - 14) = \dfrac{1}{7x - 14} * 7\)

Why the difference? It seems that \(\displaystyle \ln(cos2x)\) is differentiated twice and this is multiplied to the original problem. While, in the other one the source problem is differentiated once and this is multiplied to the source problem.

The number of differentiations by the chain rule is equal to the layers of functions.
The first example is the logarithm of the cosine of a product, hence three layers
The second is the logarithm of a polynomial - only two layers of chaining

 

[Jason76]

The number of differentiations by the chain rule is equal to the layers of functions.
The first example is the logarithm of the cosine of a product, hence three layers
The second is the logarithm of a polynomial - only two layers of chaining

Sounds good. Makes sense. Thanks.