2.12 Logarithmic Differentiation

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We use the logarithm to compute the derivative of a function.

Logarithmic Differentiation

In this section we use the natural logarithm and its properties to help us to compute the derivative of some special functions as well as some complicated functions. From the chain rule, we have $\frac {d}{dx}\left [\ln (g(x))\right ] = \frac {g'(x)}{g(x)}.$ If we solve this equation for $g'(x)$ we get the logarithmic differentiation formula: $g'(x)= g(x)\cdot \frac {d}{dx}\left [\ln (g(x))\right ].$ It says that we can find the derivative of a function by multiplying the original function by the derivative of its logarithm. The reason that this formula is potentially useful is that the logarithm function has three calculus friendly properties:

$\begin{align*} 1.\;\; &\ln(uv) = \ln(u) + \ln(v)\\ 2.\;\; &\ln\left(\frac{u}{v}\right) = \ln(u) - \ln(v)\\ 3.\;\; &\ln(u^n) = n\ln(u). \end{align*}$

In these pre-calculus formulas, the expressions $u, v$ and $n$ can be functions as well as constants.

Use logarithmic differentiation to find the derivative of $g(x) = x^x.$ We begin by computing the logarithm of $g(x)$ using the properties of the logarithm. $\ln (g(x)) = \ln (x^x) = x\ln (x).$ Now we can use the product rule to compute the derivative of $\ln (g(x))$ . We have $\frac {d}{dx} \left [\ln (g(x))\right ] = \frac {d}{dx}\left [ x\ln (x)\right ]$ $= 1\cdot \ln (x) + x \cdot \frac {1}{x} = \ln (x) + 1 = 1+ \ln (x).$ Finally, we use the logarithmic differentiation formula to obtain $g'(x)$ . We have: $g'(x) = g(x) \frac {d}{dx} \left [\ln (g(x))\right ] = x^x [1+\ln (x)].$

Compute $\frac {d}{dx} \left (x^{2x}\right )$

Use logarithmic differentiation

Use property 3 of logarithms

Use the product rule to find the derivative of $\ln (g(x))$

Remember to multiply by the original function

The derivative of $x^{2x}$ with respect to $x$ is $\answer [given]{x^{2x} (2+2\ln (x))}$

Use logarithmic differentiation to find the derivative of $g(x) = x^{\sin (x)}.$ We begin by computing the logarithm of $g(x)$ using the properties of the logarithm. $\ln (g(x)) = \ln \left (x^{\sin (x)}\right ) = \sin (x)\ln (x).$ Now we can use the product rule to compute the derivative of $\ln (g(x))$ . We have $\frac {d}{dx} \left [\ln (g(x))\right ] = \frac {d}{dx} \left [\sin (x)\ln (x)\right ]$ $= \cos (x)\cdot \ln (x) + \sin (x) \cdot \frac {1}{x}.$ Finally, we use the logarithmic differentiation formula to obtain $g'(x)$ . We have: $g'(x) = g(x) \frac {d}{dx} \left [\ln (g(x))\right ] = x^{\sin (x)} \left [\cos (x)\ln (x) + \frac {\sin (x)}{x}\right ].$

Compute $\frac {d}{dx} \left (x^{\tan (x)}\right )$

Use logarithmic differentiation

Use property 3 of logarithms

Use the product rule to find the derivative of $\ln (g(x))$

Remember to multiply by the original function

The derivative of $x^{\tan (x)}$ with respect to $x$ is $\answer [given]{x^{\tan (x)} (\sec ^2(x)\ln (x) + \frac {\tan (x)}{x})}$

Use logarithmic differentiation to find the derivative of $g(x) = \frac {e^x \sin ^4(x)}{\sqrt {x^4 + x^2 + 1}}.$ We begin by computing the logarithm of $g(x)$ using the properties of the logarithm. $\begin{align*} \ln(g(x)) &= \ln\left(\frac{e^x \sin^4(x)}{\sqrt{x^4 + x^2 + 1}}\right)\\ &= \ln\left(e^x \sin^4(x)\right) - \ln\left(\sqrt{x^4 + x^2 + 1}\right)\\ &= \ln(e^x) + \ln(\sin^4(x)) - \ln[(x^4 + x^2 + 1)^{1/2}]\\ &= x + 4\ln(\sin(x)) - \frac12 \ln(x^4 + x^2 + 1). \end{align*}$

Note that we used the fact that the logarithm and the exponential functions are inverses in the last line to write $\ln (e^x) = x$ . Now we can use the chain rule to compute the derivative of $\ln (g(x))$ . We have

$\begin{align*} \frac{d}{dx}\left[ \ln(g(x))\right] &= \frac{d}{dx}\left[ x + 4\ln(\sin(x)) - \frac12 \cdot \ln(x^4 + x^2 + 1)\right]\\ &= 1+ 4 \cdot \frac{\cos(x)}{\sin(x)} - \frac12 \cdot \frac{4x^3 + 2x}{x^4 + x^2 + 1}. \end{align*}$

Finally, we use the logarithmic differentiation formula to obtain $g'(x)$ . With a couple of minor simplifications, we have: $g'(x) = \frac {e^x \sin ^4(x)}{\sqrt {x^4 + x^2 + 1}} \left [1+ 4 \cot (x) - \frac {2x^3 + x}{x^4 + x^2 + 1}\right ].$

If $g(x) = \frac {\sqrt [4]{x^8 + x^4 + 2}}{e^{2x} \sec ^2(x)}$ then which of the following is equal to $\ln (g(x))$ ?

$\frac 14 \cdot \ln (x^8 + x^4 + 2) - 2x - 2\ln (\sec (x))$ $\frac 14 \cdot \ln (x^8 + x^4 + 2) - 2x + 2\ln (\sec (x))$ $\frac 14 \cdot \ln (x^8 + x^4 + 2) - 2x\ln (\sec (x))$

Which of the following is equal to $\frac {d}{dx} \frac {\sqrt [4]{x^8 + x^4 + 2}}{e^{2x} \sec ^2(x)}?$

$\frac {2x^7 + x^3}{x^8+x^4 +2} -2 -2 \tan (x)$ $\frac {\sqrt [4]{x^8 + x^4 + 2}}{e^{2x} \sec ^2(x)} \cdot \left [\frac {2x^7 + x^3}{x^8+x^4 +2} -2 -2 \tan (x)\right ]$ $\frac {\sqrt [4]{x^8 + x^4 + 2}}{e^{2x} \sec ^2(x)} \cdot \left [\frac {2x^7 + x^3}{x^8+x^4 +2} -2 -2 \sec (x)\right ]$

As a final note, even though $\ln (x)$ is only defined for $x>0$ , the logarithmic differentiation formula given above will yield the correct result even in situations where $g(x) <0$ . This is because $\frac {d}{dx} \ln |x| = \frac {1}{x},$ and so, by the chain rule, $\frac {d}{dx} \ln |g(x)| = \frac {g'(x)}{g(x)}.$