logarithmic differentiation-

Calculus Level 3

Consider a function defined as $h(x) =\cos^{x}x$ , this function is of the form $h(x)=f(x)^{g(x)}$ and we can differentiate this function using logarithms, However is it possible to differentiate such function using any other method?

Using logarithms to differentiate: suppose a function of maps $x$ to $y$ as follows, $y=f(x)^{g(x)}$ then $\ln(y)=g(x)\ln(f(x))$ , differentiating the equation on each side:

$\frac{y'}{y}=g'(x)\ln(f(x)) +\frac{g(x)f'(x)}{f(x)}$

There is no other way Product rule Chain rule

1 solution

Amal Hari
Oct 17, 2019

We can re-write the function as $h(x)=(e^{\ln(f(x)})^{g(x)}$

which becomes $e^{g(x)\ln(f(x))}$

We can use chain rule here now , substitute $g(x)\ln(f(x))=u$

then function becomes $h(x)=e^{u}$

$\dfrac{d(h(x))}{dx}=\dfrac{d(e^{u})}{du} \times \dfrac{du}{dx}$

$\dfrac{d(h(x))}{dx}=e^{u}\times \dfrac{du}{dx}$

$\dfrac{du}{dx} =g'(x)\ln{f(x)} + \dfrac{g(x) f'(x)}{f(x)}$

$h'(x)=h(x) \times (g'(x)\ln{f(x)} + \dfrac{g(x) f'(x)}{f(x)})$

Well, the derivative is computed using a combination of the chain and product rule. The first operation is manipulating the function to the required form and then applying the chain rule. I chose the chain rule and it turned out to be the correct option. What enables us to differentiate this function in any other way is by manipulating it to another form as you have shown. I reiterate that the process of differentiation includes the use of the product as well as the chain rule. So there are multiple correct answers to this in my view.

Karan Chatrath - 1 year, 7 months ago

Yes but the product rule is required depending on the substitution u, generally we can write it as $\dfrac{d e^{u}} {du} \times \dfrac{du}{dx}$

Amal Hari - 1 year, 7 months ago

logarithmic differentiation-

1 solution

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