Question

differentiation of y=loge x^2

Answer 1

you can see your answer

Answer 2

Hey there!

Given logarithmic value to be put under derivative is:  $\bf{y = \log_e{x^2}}$;

By applying the log rule that is;

$\bf{\log_a(b) = \dfrac{\In(b)}{\In(a)}}$

Therefore,

$\bf{= \dfrac{d}{dx} (\dfrac{\In(x^2)}{In(e)})}$

Taking the constant out of the brackets that is,

$\bf{Here \quad (a \times f)' = a \times f'}$

$\bf{= 1 \times \dfrac{d}{dx} (\In(x^2))}$

Now, by applying the chain rule in this that is;

$\bf{\dfrac{df(u)}{dx} = \dfrac{df}{du] \times \dfrac{du}{dx}}$

Here,   f = In(u),   u = x^2.

$\bf{\therefore \quad 1 \times \dfrac{d}{du} (In(u)) \dfrac{d}{dx} (x^2)}$

Now, apply the common derivative rule on this.

$\bf{= 1 \times \dfrac{1}{u} \times \dfrac{d}{dx} (x^2)}$

Now, apply the power rule in this to get rid of all the differentiation derivatives that is,

$\bf{\dfrac{d}{dx} (x^a) = a \times x^{a - 1}}$

$\bf{\therefore \quad = 1 \times \dfrac{1}{u} \times 2x^{2 - 1}}$

$\bf{\therefore \quad = 1 \times \dfrac{1}{u} \times 2x}$

Substitute the original value of variable "u" that is  u = x^2  into this newly formed equation.

$\bf{\therefore \quad = 1 \times \dfrac{1}{x^2} \times 2x}$

$\bf{= 1 \times \dfrac{1 \times 2x}{x^2}}$

$\bf{\therefore \quad = 1 \times \dfrac{2x}{x^2}}$

$\boxed{\bf{\underline{\therefore \quad Final \: \: Answer = \dfrac{2}{x}}}}$

Which is the required solution for this type of query.

[FOR GRAPH, CHECK THE ATTACHMENT]

Hope this helps you and clears the doubts for this differentiation query!!!!