Math · Calculus I · Worked example
Differentiate eˣ, ln x and sin x terms
Differentiate f(x) = 3eˣ − 2 ln x + 5 sin x, then find the slope of its graph at x = 1 (x in radians).
State the domain
ln x is defined only for x > 0, so f and its derivative live on x > 0. The sine term uses radians.
Differentiate term by term
Constant multiples stay in front. The derivative of eˣ is eˣ, of ln x is 1/x, and of sin x is cos x.
Evaluate at x = 1
Substitute x = 1: 3e¹ − 2/1 + 5 cos 1. Written with the e term last, that is 5 cos 1 − 2 + 3e. With e ≈ 2.71828 and cos 1 ≈ 0.54030, the slope is about 8.856.
Interpret the slope
At x = 1 the graph rises about 8.86 units per unit of x. The exponential term contributes most of that, 3e ≈ 8.15, and it grows as x increases while the logarithm’s contribution −2/x shrinks.
Result
f′(x) = 3eˣ − 2/x + 5 cos x for x > 0, and f′(1) = 5 cos 1 − 2 + 3e ≈ 8.856.
Your turn
Find the slope of y = eˣ + ln x at x = 1.
Show the answer and explanation
The slope is e + 1 ≈ 3.718.
dy/dx = eˣ + 1/x. At x = 1 this is e + 1 ≈ 3.718.
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