Answer:
f'(x) = 2[3tan²(x)sec²(x) - 10csc⁴(x)cot(x)]
Step-by-step explanation:
f' of tan(x) = sec²(x)
f' of csc(x) = -csc(x)cot(x)
General Power Rule: uⁿ = xuⁿ⁻¹ · u'
Step 1: Write equation
2tan³(x) + 5csc⁴(x)
Step 2: Rewrite
2(tan(x))³ + 5(csc(x))⁴
Step 3: Find derivative
d/dx 2(tan(x))³ + 5(csc(x))⁴
- General Power Rule: 2 · 3(tan(x))² · sec²(x) + 5 · 4(csc(x))³ · -csc(x)cot(x)
- Multiply: 6(tan(x))²sec²(x) - 20(csc(x))³csc(x)cot(x)
- Simplify: 6tan²(x)sec²(x) - 20csc⁴(x)cot(x)
- Factor: 2[3tan²(x)sec²(x) - 10csc⁴(x)cot(x)]
A white an expression that represents the perimeter
Step-by-step explanation:
hi
Answer:
(4,5)
Step-by-step explanation:
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The average rate of change over a given interval is the slope of the secant line through the endpoints of the interval.
$203/unit.The instantaneous rate of change at a certain point is the slope of the tangent line.
We differentiate C to get 2x. We substitute 100 for x to get that the instantaneous rate of change at 100 is
$200/unit.