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Vocabulary and core calculus differentiation methods covering problems 3.5.48, EXTRA 3.59, 3.9.85, 3.7.66, and 3.10.32 from the Pearson MyLab homework assignments.
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Exercise 3.5.48 Derivative Problem
The problem requiring the evaluation of the derivative dtdy for the trigonometric function y=3+3sec(t)8tan(t), which evaluates to dtdy=3(1+sec(t))8sec(t)=3(1+cos(t))8.
Quotient Rule
The differentiation identity applied to evaluate derivatives of ratios like y=3+3sec(t)8tan(t), stating that dtd[v(t)u(t)]=[v(t)]2u′(t)v(t)−u(t)v′(t).
Exercise EXTRA 3.59 Implicit Relation
The equation yx+y=12, which requires evaluating and simplifying y′ using implicit differentiation to obtain y′=−2x+3yy.
Implicit Differentiation
A technique applied in Exercise EXTRA 3.59 where both sides of an equation in x and y are differentiated with respect to x, treating y as a differentiable function of x, and algebraically isolating y′.
Exercise 3.9.85 Differentiation Problem
The problem requiring the derivative dxd[(1+x3)x], which evaluates via logarithmic differentiation to (1+x3)x[ln(1+x3)−x+33].
Logarithmic Differentiation
The method specified in Exercise 3.9.85 for differentiating variable-base and variable-exponent expressions y=[f(x)]g(x) by taking the natural logarithm of both sides, expanding using logarithm rules, and differentiating implicitly.
Exercise 3.7.66 Derivative Problem
The problem requiring the derivative dxdy of y=tan(15xex), which requires applying the chain rule combined with the product rule to yield dxdy=15ex(1+x)sec2(15xex).
Chain Rule
The formula used to compute the derivative of a composite function y=f(g(x)), given by dxdy=f′(g(x))⋅g′(x), applied in functions such as tan(15xex) and ln(tan(15t3)).
Exercise 3.10.32 Derivative Problem
The problem of evaluating the derivative f′(t) of the logarithmic-trigonometric composite function f(t)=ln(tan(15t3)), which evaluates to f′(t)=tan(15t3)45t2sec2(15t3).
Derivative of the Natural Logarithm
The calculus rule stating that dud[ln(u)]=u1, which combines with the chain rule to give dtd[ln(u(t))]=u(t)u′(t), as used in Exercise 3.10.32.