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