Logarithmic, parametric and second order derivatives: Continuity and Differentiability

4.1 Logarithmic differentiation

When a function has a variable in the exponent, a variable in the base, or a long product of powers, take the logarithm of both sides first. The product becomes a sum, the power becomes a multiplier, and the derivative becomes ordinary implicit differentiation. This technique handles expressions such as x to the power x and (tan x) to the power x in two lines.

The method is mechanical. Write log y equal to the logarithm of the expression, differentiate both sides with respect to x using the chain rule on the left, then multiply by y to finish. Keeping the log step visible is important because the exam awards it separately.

4.1.1 Derivatives of functions in parametric form

When x and y are both given in terms of a parameter t, the chain rule gives dy by dx as dy by dt divided by dx by dt. Differentiate each coordinate with respect to t, then divide. The answer may still contain t, which is normal.

For the second derivative, differentiate dy by dx with respect to t and divide once more by dx by dt. A common error is to differentiate dy by dx with respect to x directly, which is wrong for parametric curves.

4.1.2 Second order derivatives

A second derivative measures how the slope changes. For an implicit relation, differentiate once to get dy by dx, then differentiate again and substitute the first derivative back. For a parametric curve, use the formula above.

4.1.3 Exercise 5.5 to 5.7 coverage

Exercise 5.5 is logarithmic differentiation. Exercise 5.6 is parametric curves, and Exercise 5.7 is second order derivatives. The Miscellaneous Exercise mixes all three with the earlier continuity and differentiability material. Working one item from each exercise type is the fastest revision before the exam.

4.2 Three worked derivatives

First, y equal to x to the power x. Taking logs gives log y equal to x log x. Differentiating both sides gives y' divided by y equal to log x plus 1, so y' equals x to the power x times (log x plus 1). The logarithm step is what makes the power rule legal here.

Second, y equal to cos cubed of sec squared 2t. Chain the three layers: the cube gives 3 times the square, the cosine gives minus sine of the inner expression, and the secant squared gives 4 sec squared 2t tan 2t. Multiplying the three pieces gives the derivative with respect to t.

Third, x equal to e to the power sin inverse t and y equal to e to the power cos inverse t. Then dx by dt is e to the power sin inverse t over the square root of one minus t squared, and dy by dt is minus e to the power cos inverse t over the same square root. Dividing gives dy by dx equal to minus y by x, which simplifies nicely at any particular t.

4.3 Presentation habits

Write the given function, then one line naming the rule, then the derivative. For logarithmic differentiation, show the log step separately. For parametric problems, write dx by dt and dy by dt on separate lines before dividing. For second derivatives, substitute the first derivative before differentiating again. These habits match the marking scheme and reduce sign errors.

4.4 More logarithmic differentiation

For y equal to (sin x) to the power cos x, taking logs gives log y equal to cos x log(sin x). Differentiating gives y prime over y equal to minus sin x log(sin x) plus cos squared x over sin x. Multiplying by y finishes the answer, and the same pattern handles every variable power.

For a product of several powers the logarithm turns the product into a sum, and the derivative becomes a sum of fractions. That is usually shorter than applying the product rule repeatedly.

4.5 Second derivatives of implicit relations

Differentiate the relation once and solve for y prime. Differentiate the resulting equation again, remembering that y prime is a function of x, then substitute the first derivative. The expression usually simplifies after substitution, and writing that step makes the working clear.

4.6 Exercise 5.5 to 5.7 revision order

Do one logarithmic item, one parametric item and one second derivative item from each cluster, then attempt the Miscellaneous Exercise. The paper pattern is predictable: one continuity constant, one differentiation, one logarithmic or parametric derivative and one second derivative.

4.7 Second derivative of an implicit relation

For x squared plus y squared equal to 25, the first derivative is minus x over y. Differentiating again gives minus (y plus x times y prime) over y squared, and substituting y prime gives minus (y squared plus x squared) over y cubed. Using the original relation turns the numerator into minus 25, so the second derivative is minus 25 over y cubed.