Derivative of f(x)=6\sqrt(x)+5cos\theta ?

1 Answer
Feb 13, 2018

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\qquad \qquad \qquad \qquad \qquad \qquad \quad f'(x) \ = \ 3 / \sqrt{x} - 5 sin( x ).

Explanation:

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"As" \ x \ "is the variable of" \ f(x), "I take it that the variable meant in" \ cos \ "is" \ x, "not" \ \theta. "If this is the case, we proceed as follows:"

"We are given:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ 6 \sqrt{x} + 5 cos( x ).

"First rewrite" \ f(x):

\qquad \qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ 6 x^{1/2} + 5 cos( x ).

"Now use the Sum Rule, and then the rules for the basic functions"
"present there:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad f'(x) \ = \ [ 6 x^{1/2} ]' + [ 5 cos( x ) ]'

\qquad \qquad \qquad \qquad \qquad \qquad \qquad f'(x) \ = \ 6 [ x^{1/2} ]' + 5 [ cos( x ) ]'

\qquad \qquad \qquad \qquad \qquad \qquad \qquad f'(x) \ = \ 6 [ 1/2 x^{- 1/2} ] + 5 ( -sin( x ) )

\qquad \qquad \qquad \qquad \qquad \qquad \qquad f'(x) \ = \ 6 \cdot 1/2 x^{- 1/2} - 5 sin( x )

\qquad \qquad \qquad \qquad \qquad \qquad \qquad f'(x) \ = \ 3 x^{- 1/2} - 5 sin( x ).

"Now remove the negative exponents, and write the fractional"
"exponent as a radical here:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad f'(x) \ = \ 3 ( 1/ x^{ 1/2} ) - 5 sin( x )

\qquad \qquad \quad :. \qquad \qquad \qquad f'(x) \ = \ 3 / \sqrt{x} - 5 sin( x ).

"This is our answer."

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"Summarizing:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ 6 \sqrt{x} + 5 cos( x ).

\qquad \qquad \qquad \qquad \qquad \qquad \quad f'(x) \ = \ 3 / \sqrt{x} - 5 sin( x ).