Question

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

Answer 1

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
$\dfrac{dy}{dx}=\dfrac{1}{4x-{{x}^{5}}}\dfrac{d}{dx}(4x-{{x}^{5}})$

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,
\[\dfrac{dy}{dx}=\dfrac{1}{4x-{{x}^{5}}}(4-5{{x}^{4}})\]
By further simplification we can write,
$\dfrac{dy}{dx}=\dfrac{4-5{{x}^{4}}}{4x-{{x}^{5}}}$
Hence, the answer is $\dfrac{4-5{{x}^{4}}}{4x-{{x}^{5}}}$

Note:
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,
$4x-{{x}^{5}}={{e}^{y}}$
Again we can differentiate with respect to x both sides and proceed.
$\dfrac{d}{dx}(4x-{{x}^{5}})=\dfrac{d}{dx}({{e}^{y}})$
Now we can apply chain rule and proceed as earlier,
$(4-5{{x}^{4}})={{e}^{y}}\dfrac{dy}{dx}$
Dividing both sides with ${{e}^{y}}$ we have,
$\dfrac{dy}{dx}=\dfrac{4-5{{x}^{4}}}{{{e}^{y}}}$
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
$\dfrac{dy}{dx}=\dfrac{4-5{{x}^{4}}}{4x-{{x}^{5}}}$
As we can see by both methods we get exactly the same answer.