Mathematics, 29.08.2019 01:50 tishfaco5000

Chain rule : calculus problem
find the derivative of
f(x) = sqrt[cos(5x)]
the answer is supposedly -5sin(5x)/2sqrtcos(5x)
how does one arrive at this answer?

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Answer from: ashlynneboogs0056

$\displaystyle f'(x) = \frac{-5sin(5x)}{2\sqrt{cos(5x)}}$

General Formulas and Concepts:

Algebra I

Functions

Function Notation

Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

DerivativesDerivative Notation

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

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

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

Trig Derivatives

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \sqrt{cos(5x)}$

Step 2: Differentiate

Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle f(x) = [cos(5x)]^\bigg{\frac{1}{2}}$ Derivative Rule [Chain Rule]: $\displaystyle f'(x) = \frac{d}{dx} \bigg[ [cos(5x)]^\bigg{\frac{1}{2}} \bigg] \cdot \frac{d}{dx}[cos(5x)] \cdot \frac{d}{dx}[5x]$ Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle f'(x) = \frac{d}{dx} \bigg[ [cos(5x)]^\bigg{\frac{1}{2}} \bigg] \cdot \frac{d}{dx}[cos(5x)] \cdot 5\frac{d}{dx}[x]$ Basic Power Rule: $\displaystyle f'(x) = \frac{1}{2}[cos(5x)]^\bigg{\frac{1}{2} - 1} \cdot \frac{d}{dx}[cos(5x)] \cdot 5x^{1 - 1}$ Simplify: $\displaystyle f'(x) = \frac{5}{2}[cos(5x)]^\bigg{\frac{-1}{2}} \cdot \frac{d}{dx}[cos(5x)]$ Trig Derivative: $\displaystyle f'(x) = \frac{5}{2}[cos(5x)]^\bigg{\frac{-1}{2}} \cdot -sin(5x)$ Simplify: $\displaystyle f'(x) = \frac{-5sin(5x)}{2}[cos(5x)]^\bigg{\frac{-1}{2}}$ Rewrite [Exponential Rule - Rewrite]: $\displaystyle f'(x) = \frac{-5sin(5x)}{2[cos(5x)]^\bigg{\frac{1}{2}}}$ Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle f'(x) = \frac{-5sin(5x)}{2\sqrt{cos(5x)}}$

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

Unit: Derivatives

Book: College Calculus 10e

Answer from: Quest

what?

step-by-step explanation:

Answer from: Quest

none of the above.

for one of the equations to be used in the process, there has to be a $2x^2$ term in the equation since $2x$ squared is $2x^2$ .

since there is only a $x^2$ term in the first equation, we cannot derive a $2x+4$ from completing the square.

solving with completing the square

$x^2=-4x+3$

$x^2+4x=3$

$x^2+4x+4=7$

$(x+2)^2=7$

$x+2=+-\sqrt{7}$

$x=2+-\sqrt{7}$

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Chain rule : calculus problem
find the derivative of
f(x) = sqrt[cos(5x)]
the an...

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