Apply logarithmic derivatives

At P, what is the first calculation for the tangent gradient?

Logarithmic tangent and normal lines

The derivative of a logarithmic function gives the gradient of its tangent. Use the point and that gradient for a tangent, or use its negative reciprocal for a normal

1. 1Differentiate the logarithm For a logarithm of an inner function, divide the inner derivative by the inner function \(\displaystyle \frac{d}{dx}\log_e\left(g(x)\right)=\frac{g'(x)}{g(x)}\)
2. 2Evaluate at the point Substitute the given x-value into the derivative to obtain the tangent gradient \(\displaystyle f(x)=2\log_e\left(x+1\right)+3\quad\Rightarrow\quad f'(0)=\frac{2}{0+1}=2\)
3. 3Use point-slope form A tangent line uses the point on the curve and the gradient from the derivative \(\displaystyle m=2,\quad P=(0,3)\quad\Rightarrow\quad y-3=2\left(x-0\right)\)
4. 4Use a negative reciprocal for a normal For a non-zero tangent gradient, the normal gradient is its negative reciprocal \(\displaystyle m=2\quad\Rightarrow\quad m_{n}=-\frac{1}{2}\)

Find the tangent gradient of f at x=0

Find the y-intercept of the tangent through P with the given gradient

Which tangent gradient is correct?

A tangent at an x-intercept

First find where the original function is zero. For this function, the logarithm argument equals 1 there, not zero. Then differentiate and use the intercept point

1. 1Find the intercept The logarithm vanishes when its argument is 1 \(\displaystyle F(x)=2\log_e(3x-2),\quad 3x-2=1\Rightarrow x=1\)
2. 2Find the tangent The intercept lies inside the original domain x > 2/3 \(\displaystyle F'(x)=\frac{6}{3x-2},\quad F'(1)=6,\quad y=6(x-1)=6x-6\)

Find the x-coordinate of the x-intercept. Enter an exact number

Use the normal gradient

The normal is perpendicular to the tangent at the same point. Use the negative reciprocal of a non-zero tangent gradient. A horizontal tangent instead has a vertical normal

1. 1Evaluate the tangent gradient At x = 0, the point on the curve is (0,3) \(\displaystyle F(x)=2\log_e(x+1)+3,\quad F'(0)=2\)
2. 2Form the normal Keep the same point but change the gradient \(\displaystyle m_n=-\frac12,\quad y-3=-\frac12(x-0)\)

Find the normal gradient at x = 0. Enter an exact value

Identify the line in the condition

A tangent parallel to a reference line has the reference gradient. A normal perpendicular to that reference line has its negative reciprocal. These can give the same contact point, but they request different lines

1. 1Tangent parallel to the reference Use equal tangent and reference gradients \(\displaystyle F(x)=\log_e(2x+1),\quad y_{\mathrm{ref}}=2x-3,\quad F'(x)=2\Rightarrow x=0\)
2. 2Normal perpendicular to the reference Taking negative reciprocals on both sides gives the equivalent tangent-gradient condition F′(x) = 2 \(\displaystyle -\frac{1}{F'(x)}=-\frac12\Rightarrow F'(x)=2\Rightarrow x=0\)
3. 3Write the requested line Both lines pass through (0,0); their gradients are different \(\displaystyle y_{\mathrm{tangent}}=2x,\quad y_{\mathrm{normal}}=-\frac12x\)

The normal is perpendicular to the reference line. Find the x-coordinate of contact

Curve

\(\displaystyle F(x)=-2\log_e(3x-2)+1\)

For this curve at x = 1, the normal gradient is 1/6. Find the normal's y-intercept

A stationary point with exact coordinates

Differentiate the product, solve F′(x) = 0 in x > 0, then substitute into F. The sign of F′′ classifies the stationary point

1. 1Differentiate the product The derivative of x log_e(x) is log_e(x) + 1 \(\displaystyle F(x)=2x\log_e(x)-4x+3,\quad F'(x)=2\log_e(x)-2\)
2. 2Solve and substitute The inverse of the natural logarithm gives an exact exponential coordinate \(\displaystyle \log_e(x)=1\Rightarrow x=e,\quad F(e)=3-2e\)
3. 3Classify the point Since x > 0, F′′ is positive, so the point is a local minimum \(\displaystyle F''(x)=\frac2x>0\)
4. 4Express the coordinates in the given form The coordinates have the following form and coefficients \(\displaystyle (x,y)=(e,3-2e),\quad x=Ue^P,\ y=Ve^Q+W,\quad U=P=Q=1,\ V=-2,\ W=3\)

Find P when the stationary x-coordinate is written as x = e^P. Enter a number

Classify the stationary point of G on x > 0

The stationary point has x = e². Find V when its y-coordinate is Ve² - 2