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4 years ago

Derivative of 2 tan^3(x)+5 csc^4(x) ?

Mathematics

1 answer:

Leni [432]4 years ago

3 0

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)]

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A. There is a focus at (0,−10).

Step-by-step explanation:

Assume the hyperbola is like the one below.

The hyperbola is vertical and centred on the y-axis, so its general equation is

$\dfrac{y^{2}}{a^{2}} - \dfrac{x^{2}}{b^{2}} = 1$

The vertices of your parabola are (0,±8) so a = 8.

The covertices are (±6,0), so b = 6.

Calculate c

$\begin{array}{rcl}a^{2} + b^{2} & = & c^{2}\\8^{2} + 6^{2} & = & c^{2}\\64 + 36 & = & c^{2}\\100 & = & c^{2}\\c & = & \mathbf{10}\\\end{array}$

A. Foci

The foci are at (0, ±c) = (0, ±10)

TRUE. There is a focus at (0,-10).

B. Foci

The foci are at (0,±10).

False. There is no focus at (0,12)

C. and D. Asymptotes

The equations for the asymptotes are

$y = \pm\dfrac{a}{b}x = \pm\dfrac{8}{6}x = \pm\dfrac{4}{3}x$

So, y = ±x are not asymptotes.

False.

E. and F. Directrices

The equations of the directrices are

y = ±a²/c = ±64/10 = ±6.4

y = 6.4 is a directrix.

E is false. x = cannot be a directrix

F is uncertain. Your equation for the directrix is incomplete.

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