Question

Calculus 1: Logorithmic differentiation

Answer 1

Given,

$\longrightarrow y=(6+6x)^{\frac{4}{x}}$

At $x=1,$

$\longrightarrow y=(6+6(1))^{\frac{4}{1}}$

$\longrightarrow y=12^4$

Taking log on both sides,

$\longrightarrow \log y=\log\left[(6+6x)^{\frac{4}{x}}\right]$

Since $\log\left(a^b\right)=b\log a,$

$\longrightarrow \log y=\dfrac{4}{x}\log(6+6x)$

Differentiating wrt x,

$\longrightarrow\dfrac{d}{dx}\left(\log y\right)=\dfrac{d}{dx}\left[\dfrac{4}{x}\log(6+6x)\right]$

By chain rule,

• $\dfrac{du}{dx}=\dfrac{du}{dy}\cdot\dfrac{dy}{dx}$

and product rule,

• $\dfrac{d}{dx}(uv)=\dfrac{du}{dx}\cdot v+u\cdot\dfrac{dv}{dx}$

we get,

$\longrightarrow\dfrac{d}{dy}\left(\log y\right)\cdot\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{4}{x}\right)\log(6+6x)+\dfrac{4}{x}\cdot\dfrac{d}{dx}\left[\log(6+6x)\right]$

$\longrightarrow\dfrac{1}{y}\cdot\dfrac{dy}{dx}=-\dfrac{4}{x^2}\log(6+6x)+\dfrac{4}{x}\cdot\dfrac{6}{6+6x}$

$\longrightarrow\dfrac{1}{y}\cdot\dfrac{dy}{dx}=-\dfrac{4}{x^2}\log(6+6x)+\dfrac{4}{x(1+x)}$

$\longrightarrow\dfrac{dy}{dx}=y\left(\dfrac{4}{x(1+x)}-\dfrac{4\log(6+6x)}{x^2}\right)$

$\longrightarrow\dfrac{dy}{dx}=4y\left(\dfrac{x^2-x(1+x)\log(6+6x)}{x^3(1+x)}\right)$

$\longrightarrow\dfrac{dy}{dx}=4\left(6+6x\right)^{\frac{4}{x}}\left(\dfrac{x-(1+x)\log(6+6x)}{x^2(1+x)}\right)$

$\longrightarrow\underline{\underline{\dfrac{dy}{dx}=\dfrac{4\left(6+6x\right)^{\frac{4}{x}}\left[x-(1+x)\log(1+x)-(1+x)\log6\right]}{x^2(1+x)}}}$

The slope of the tangent line at $x=1$ is given by,

$\longrightarrow\dfrac{dy}{dx}(x=1)=\dfrac{4\left(6+6(1)\right)^{\frac{4}{1}}\left[1-(1+1)\log(1+1)-(1+1)\log6\right]}{1^2(1+1)}$

$\longrightarrow\underline{\underline{\dfrac{dy}{dx}(x=1)=12^4\left(2-4\log12\right)}}$

Then the equation of the tangent line at $x=1$ will be,

$\longrightarrow y-12^4=12^4\left(2-4\log 12\right)(x-1)$

$\longrightarrow\underline{\underline{12^4\left(2-4\log 12\right)x-y-12^4\left(1-4\log 12\right)=0}}$