Mathematics, 22.06.2019 03:30 fluffypuppies24

Find the derivative of the function

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$\displaystyle y' = \frac{9 \bigg[ 6x^\big{\frac{9}{4}} \sqrt{x^3} \sin (x^3) - \sin (\sqrt{x}) \bigg] }{2x^\big{\frac{1}{4}}}$

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 2]: $\displaystyle \frac{d}{dx}[\int\limits^x_a {f(t)} \, dt] = f(x)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Flipping Integral]: $\displaystyle \int\limits^b_a {f(x)} \, dx = -\int\limits^a_b {f(x)} \, dx$

Integration Property [Splitting Integral]: $\displaystyle \int\limits^c_a {f(x)} \, dx = \int\limits^b_a {f(x)} \, dx + \int\limits^c_b {f(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = \int\limits^{x^3}_{\sqrt{x}} {9\sqrt{t} \sin (t)} \, dt$

Step 2: Differentiate

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle y = 9\int\limits^{x^3}_{\sqrt{x}} {\sqrt{t} \sin (t)} \, dt$ [Integral] Rewrite [Integration Property - Splitting Integral]: $\displaystyle y = 9 \bigg[ \int\limits^0_{\sqrt{x}} {\sqrt{t} \sin (t)} \, dt + \int\limits^{x^3}_0 {\sqrt{t} \sin (t)} \, dt \bigg]$ [1st Integral] Rewrite [Integration Property - Flipping Integral]: $\displaystyle y = 9 \bigg[ -\int\limits^{\sqrt{x}}_0 {\sqrt{t} \sin (t)} \, dt + \int\limits^{x^3}_0 {\sqrt{t} \sin (t)} \, dt \bigg]$ Chain Rule [Integration Rule - Fundamental Theorem of Calculus 2]: $\displaystyle y' = 9 \bigg[ - \bigg( {\sqrt{\sqrt{x}} \sin (\sqrt{x}) \bigg) \cdot \frac{d}{dx}[\sqrt{x}] + \bigg( \sqrt{x^3} \sin (x^3) \bigg) \cdot \frac{d}{dx}[x^3] \bigg]$ Basic Power Rule: $\displaystyle y' = 9 \bigg[ - \bigg( {\sqrt{\sqrt{x}} \sin (\sqrt{x}) \bigg) \frac{1}{2\sqrt{x}} + \bigg( \sqrt{x^3} \sin (x^3) \bigg) \cdot 3x^2 \bigg]$ Simplify: $\displaystyle y' = 9 \bigg[ \frac{- \bigg( x^\big{\frac{1}{4}} \sin (\sqrt{x}) \bigg)}{2\sqrt{x}} + \bigg( \sqrt{x^3} \sin (x^3) \bigg) \cdot 3x^2 \bigg]$ Rewrite: $\displaystyle y' = \frac{9 \bigg[ 6x^\big{\frac{9}{4}} \sqrt{x^3} \sin (x^3) - \sin (\sqrt{x}) \bigg] }{2x^\big{\frac{1}{4}}}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Answer from: Quest

image result for what is another name for consecutive interior angles?

when two lines are crossed by another line (called the transversal): the pairs of angles on one side of the transversal but inside the two lines are called consecutive interior angles. in this example, these are consecutive interior angles: d and f.

step-by-step explanation:

Answer from: Quest

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2