As we have seen with other functions, we can use our knowledge of differentiating logarithmic functions to find the equation of tangents and normals to such functions.
$\frac{d}{dx}\ln f\left(x\right)$ddxlnf(x) | $=$= | $\frac{f'\left(x\right)}{f\left(x\right)}$f′(x)f(x) |
When finding equations of tangents and normals, remember that the gradient of the tangent at any point on a function is given by the first derivative of the function. The following formulas are also useful:
Given the gradient $m$m and a point $\left(x_1,y_1\right)$(x1,y1), the equation of a line can be found using the point gradient formula:
$y-y_1=m\left(x-x_1\right)$y−y1=m(x−x1)
Given the gradient of the tangent $m_1$m1 at a point, the gradient of the normal $m_2$m2 is:
$m_2=-\frac{1}{m_1}$m2=−1m1
Find the gradient of the tangent to the curve $y=\ln\left(4x-8\right)$y=ln(4x−8) at the point where $x=3$x=3.
Solution: Finding the derivative of $y=\ln\left(4x-8\right)$y=ln(4x−8) gives:
$\frac{dy}{dx}$dydx | $=$= | $\frac{4}{4x-8}$44x−8 |
$=$= | $\frac{1}{x-2}$1x−2 |
So at $x=3$x=3, the gradient of the tangent is:
$\frac{dy}{dx}$dydx | $=$= | $\frac{1}{3-2}$13−2 |
$=$= | $1$1 |
Consider the function $f\left(x\right)=\ln x$f(x)=lnx.
Determine the $x$x value of the $x$x-intercept.
Determine the equation of the tangent line to the curve at the point where it crosses the $x$x-axis.
Determine the equation of the normal line to the curve at the point where it crosses the $x$x-axis.
The first and second derivatives of logarithmic functions can be used to determine concavity, and regions where a logarithmic function is increasing or decreasing. Additionally, any stationary points and points of inflection can be found. Using these tools we can sketch functions involving logarithms.
Consider the graph of $y=\ln x$y=lnx.
Is the function increasing or decreasing?
Increasing
Decreasing
Is the gradient to the curve negative at any point on the curve?
No
Yes
Which of the following best completes this sentence?
"As $x$x increases, the gradient of the tangent..."
decreases at a constant rate.
increases at an increasing rate.
increases at a constant rate.
decreases at an increasing rate.
increases at a decreasing rate.
decreases at a decreasing rate.
Which of the following best completes the sentence?
"As $x$x gets closer and closer to $0$0, the gradient of the tangent..."
increases towards a fixed value.
decreases towards $-\infty$−∞.
decreases towards $0$0.
increases towards $\infty$∞.
We have found that the gradient function must be a strictly positive function, and it must also be a function that decreases at a decreasing rate. What kind of function could it be?
Quadratic, of the form $y'=ax^2$y′=ax2.
Exponential, of the form $y'=a^{-x}$y′=a−x.
Linear, of the form $y=ax$y=ax.
Hyperbolic, of the form $y'=\frac{a}{x}$y′=ax.
Consider the function $f\left(x\right)=x-\ln x$f(x)=x−lnx.
Find the values of $x$x where $f'\left(x\right)=0$f′(x)=0.
Find $f''\left(x\right)$f′′(x).
Determine the nature of the stationary point.
Minimum
Point of inflection
Maximum
Which of the following best describes the behaviour of this function?
As $x$x$\to$→$\infty$∞, $y$y$\to$→$-\infty$−∞.
As $x$x$\to$→$0$0, $y$y$\to$→$-\infty$−∞.
As $x$x$\to$→$\infty$∞, $y$y$\to$→$0$0.
As $x$x$\to$→$0$0, $y$y$\to$→$\infty$∞.
As $x$x$\to$→$\infty$∞, $y$y$\to$→$\infty$∞.
As $x$x$\to$→$0$0, $y$y$\to$→$0$0.
As $x$x$\to$→$\infty$∞, $y$y$\to$→$\infty$∞.
As $x$x$\to$→$0$0, $y$y$\to$→$\infty$∞.
Which is the correct sketch of the graph?