Calculus Differentiation Techniques and Exercises

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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.

Last updated 2:56 AM on 10/15/26
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Exercise 3.5.48 Derivative Problem

The problem requiring the evaluation of the derivative dydt\frac{dy}{dt} for the trigonometric function y=8tan⁡(t)3+3sec⁡(t)y = \frac{8\tan(t)}{3 + 3\sec(t)}, which evaluates to dydt=8sec⁡(t)3(1+sec⁡(t))=83(1+cos⁡(t))\frac{dy}{dt} = \frac{8\sec(t)}{3(1 + \sec(t))} = \frac{8}{3(1 + \cos(t))}.

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Quotient Rule

The differentiation identity applied to evaluate derivatives of ratios like y=8tan⁡(t)3+3sec⁡(t)y = \frac{8\tan(t)}{3 + 3\sec(t)}, stating that ddt[u(t)v(t)]=u′(t)v(t)−u(t)v′(t)[v(t)]2\frac{d}{dt}\left[\frac{u(t)}{v(t)}\right] = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}.

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Exercise EXTRA 3.59 Implicit Relation

The equation yx+y=12y\sqrt{x + y} = 12, which requires evaluating and simplifying y′y' using implicit differentiation to obtain y′=−y2x+3yy' = -\frac{y}{2x + 3y}.

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Implicit Differentiation

A technique applied in Exercise EXTRA 3.59 where both sides of an equation in xx and yy are differentiated with respect to xx, treating yy as a differentiable function of xx, and algebraically isolating y′y'.

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Exercise 3.9.85 Differentiation Problem

The problem requiring the derivative ddx[(1+3x)x]\frac{d}{dx}\left[\left(1 + \frac{3}{x}\right)^x\right], which evaluates via logarithmic differentiation to (1+3x)x[ln⁡(1+3x)−3x+3]\left(1 + \frac{3}{x}\right)^x \left[\ln\left(1 + \frac{3}{x}\right) - \frac{3}{x + 3}\right].

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Logarithmic Differentiation

The method specified in Exercise 3.9.85 for differentiating variable-base and variable-exponent expressions y=[f(x)]g(x)y = [f(x)]^{g(x)} by taking the natural logarithm of both sides, expanding using logarithm rules, and differentiating implicitly.

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Exercise 3.7.66 Derivative Problem

The problem requiring the derivative dydx\frac{dy}{dx} of y=tan⁡(15xex)y = \tan(15xe^x), which requires applying the chain rule combined with the product rule to yield dydx=15ex(1+x)sec⁡2(15xex)\frac{dy}{dx} = 15e^x(1 + x)\sec^2(15xe^x).

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Chain Rule

The formula used to compute the derivative of a composite function y=f(g(x))y = f(g(x)), given by dydx=f′(g(x))⋅g′(x)\frac{dy}{dx} = f'(g(x)) \cdot g'(x), applied in functions such as tan⁡(15xex)\tan(15xe^x) and ln⁡(tan⁡(15t3))\ln(\tan(15t^3)).

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Exercise 3.10.32 Derivative Problem

The problem of evaluating the derivative f′(t)f'(t) of the logarithmic-trigonometric composite function f(t)=ln⁡(tan⁡(15t3))f(t) = \ln(\tan(15t^3)), which evaluates to f′(t)=45t2sec⁡2(15t3)tan⁡(15t3)f'(t) = \frac{45t^2\sec^2(15t^3)}{\tan(15t^3)}.

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Derivative of the Natural Logarithm

The calculus rule stating that ddu[ln⁡(u)]=1u\frac{d}{du}[\ln(u)] = \frac{1}{u}, which combines with the chain rule to give ddt[ln⁡(u(t))]=u′(t)u(t)\frac{d}{dt}[\ln(u(t))] = \frac{u'(t)}{u(t)}, as used in Exercise 3.10.32.