Find $\dfrac{dy}{dx}$ , if $y=\log (4x-{{x}^{5}})$ .

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Hint: To solve this question we should use chain rule and also remember differentiation of $\log x$ .

Complete step-by-step answer:
We will need following facts to solve this question
(a) Differentiation of ${{\log }_{e}}x$ is given by- $\dfrac{d}{dx}\left( {{\log }_{e}}x \right)=\dfrac{1}{x}$ ……….(i)
(b) Differentiation of $k{{x}^{n}}$ where k and n are constants is given by- $\dfrac{d}{dx}(k{{x}^{n}})=k.n.{{x}^{n-1}}$ ………(ii)

We have $y=\log (4x-{{x}^{5}})$ . Differentiating both sides with respect to x we get,
$\dfrac{dy}{dx}=\dfrac{d}{dx}\{\log (4x-{{x}^{5}})\}$

Now we will proceed step by step.
We differentiate first the logarithmic term with the help of equation (i) according to the chain rule

Now we need to differentiate the polynomial term. For this we write,
\[\dfrac{dy}{dx}=\dfrac{1}{4x-{{x}^{5}}}\left\{ \dfrac{d}{dx}(4x)-\dfrac{d}{dx}({{x}^{5}}) \right\}\]
Now we use equation (ii) to solve this differential. So, we have,
By further simplification we can write,
Hence, the answer is $\dfrac{4-5{{x}^{4}}}{4x-{{x}^{5}}}$

We should also keep in mind the base of the logarithm when we are differentiating. We must make the base of logarithm as e when we try to differentiate as equation (i) is only valid when the base is e.
There can be another way to solve this differentiation.
As we know,
${{\log }_{a}}b=c\Rightarrow b={{a}^{c}}$
We can use the following result too in our question and proceed.
We have, $y=\log (4x-{{x}^{5}})$
Using the above property we can write,
Again we can differentiate with respect to x both sides and proceed.
Now we can apply chain rule and proceed as earlier,
Dividing both sides with ${{e}^{y}}$ we have,
Substituting y we have,
$\dfrac{dy}{dx}=\dfrac{4-5{{x}^{4}}}{{{e}^{\log (4x-{{x}^{5}})}}}$
Now we need the following property of exponential: ${{a}^{{{\log }_{a}}b}}=b$
Therefore, we can write
As we can see by both methods we get exactly the same answer.
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