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Differentiation test bite

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Differentiation - test bite answers

Question

Carry out the following differentiations:

a) ${d \over {dx}}\left( {x^2 } \right)$

b) ${d \over {dx}}\left( {2x^3 } \right)$

c) ${d \over {dx}}\left( {x^{ - 4} } \right)$

d) ${d \over {dx}}\left( {x^{{1 \over 3}} } \right)$

e) ${d \over {dx}}\left( {x^{ - {1 \over 4}} } \right)$

Answer

c) $- 4x^{ - 5}$

d) ${1 \over 3}x^{{{ - 2} \over 3}}$

e) $- {1 \over 4}\;x^{ - {5 \over 4}}$

Question

Find the derived function y'(x)

a) $y = \left( {x + 1} \right)\left( {x^2 + 5} \right)$

b) $y = {{2x + 1} \over x}$

c) $y = {{3x^2 - 5} \over {\sqrt x }}$

Answer

$\eqalign{ y & = & \left( {x + 1} \right)\;\left( {x^2 + 5} \right) \cr \cr & = & x^3 + x^2 + 5x + 5 \cr \cr \Rightarrow y'\left( x \right)} & = & 3\;x^2 + 2x + 5 \cr}$

$\eqalign{ y & = & {{2x + 1} \over x} \cr \cr & = & {{2x} \over x} + {1 \over x} \cr \cr & = & 2 + x^{ - 1} \cr \cr \Rightarrow y'\left( x \right) & = & - x^{ - 2} \cr \cr & = & - {1 \over {x^2 }} \cr}$

$\eqalign{ y & = & {{3\;x^2 - 5} \over {\sqrt x }} \cr \cr & = & {{3x^2 } \over {x^{{1 \over 2}} }} - {5 \over {x^{{1 \over 2}} }} \cr \cr & = & 3x^{{3 \over 2}} - 5x^{ - {1 \over 2}} \cr \cr \Rightarrow y'\left( x \right) & = & {9 \over 2}x^{{1 \over 2}} + {5 \over 2}\;x^{ - {3 \over 2}}\cr}$

Question: Find the equation of the tangent to the curve $y = {4 \over {\sqrt x }}$ where x = 4.

Answer

$\eqalign{ y\;\left( 4 \right) & = & {4 \over {\sqrt 4 }} \cr \cr & = & 2 \cr \cr y & = & 4x^{ - {1 \over 2}} \cr \cr \Rightarrow y'\left( x \right) & = & - 2x^{ - {3 \over 2}} \cr \cr \Rightarrow y'\;(4) & = & {{ - 2} \over {4^{{3 \over 2}} }} \cr \cr & = & {{ - 2} \over 8} \cr \cr & = & \, - {1 \over 4} \cr}$

Hence the equation is $y - 2 = - {1 \over 4}\left( {x - 4} \right)$ . That is, $\eqalign{ 4y - 8 & = & - x + 4 \cr \Rightarrow x + 4y & = & 12 \cr}$

Question: At what point on the curve is the tangent parallel to ?

Answer

$2x + y = 0 \Rightarrow y = - 2x$ which has gradient -2

$\eqalign{ && y = 3x^2 + 4x + 5 \cr \cr && \Rightarrow y' = 6x + 4 \cr}$

We need

$\Rightarrow x = - 1, y = 4$ . At (-1, 4) the tangent to the curve is parallel to 2x + y = 0

Question

Is the function $f(x) = {{x^2 + 3} \over x}$ decreasing, increasing or stationary when x is equal to:

a) 1

b) $\sqrt 3$

c) 2

Answer

$\eqalign{ f\;\left( x \right) & = & {{x^2 + 3} \over x} \cr \cr & = & x + 3x^{ - 1} \cr \cr \Rightarrow f'\;(x) & = & 1 - 3x^{ - 2} \cr \cr & = & 1 - {3 \over {x^2 }} \cr}$

a) $f'\;(1) = - 2\;\;$ which is less than 0, so the function is decreasing

b) $f'\;(\sqrt 3 ) = 0$ the function is stationary

c) $f'\;(2) = {1 \over 4}$ which is greater than 0, so the function is increasing

Question: Find the stationary points on the curve and determine their nature.

Answer

$\eqalign{ &&y = x^3 + 3x^2 + 3x + 2 \cr \cr &&\Rightarrow y'(x) = 3x^2 + 6x + 3 \cr \cr && = 3\;\left( {x^2 + 2x + 1} \right) \cr}$

$= 3\;\left( {x + 1} \right)^2 = 0$ for stationary points

$\Rightarrow x = - 1$

x		-1
3(x + 1)² = y^'	+ ve	0	+ ve
tangent

When x = -1, y = 1, so the point (-1, 1) is a point of inflection. The stationary point is (-1, 1)

Question: Sketch the curve with the equation .

Answer

$\eqalign{ &&x = 0 \Rightarrow \,y = 0 \cr \cr &&\Rightarrow \left( {0,0} \right) \cr \cr &&y = 0 \cr \cr && \Rightarrow x^4 - 4x^3 = 0 \cr \cr &&x^3 (x - 4) = 0 \cr \cr && \Rightarrow x = 0,\;0,\;0,\;4 \cr \cr && \Rightarrow \left( {0,\;0} \right)\;\& \left( {4,\;0} \right) \cr \cr && y'\left( x \right) = 4x^3 - 12x^2 \cr}$

$= 4x^2 \left( {x - 3} \right) = 0$ for stationary points

$\eqalign{ \Rightarrow x & = & 0,\;0,\;3 \cr y & = & 0,\;0, - 27 \cr}$

x		0			3
4x² (x - 3) = y^'	- ve	0	- ve	- ve	0	+ ve
tangent

(falling) point of inflection at (0, 0)

minimum turning point at (3, -27)

Graph of y = x to power of four - 4x cubed

Question

The point P lies in the first quadrant on the line with equation . A rectangle is formed with sides parallel to the axes and vertices at P and the origin.

Find the position of P for which the shaded area is greatest.

Shaded area under straight line with point P

Answer

Let the point P have coordinates (x,y). So the area required, A, is given by xy. Using the equation of the line we find that

becomes $\eqalign{ A & = & x\left( {10 - 2x} \right) \cr & = & 10x - 2x^2 \cr}$

Thus $A'\;(x) = 10 - 4x = 0$ for stationary points

$\eqalign{ &\Rightarrow& x = 2.5 \cr &\Rightarrow& P\;is\;(2.5,\;5) \cr}$

We can tell this is a maximum turning point because we're dealing with a quadratic function with a negative leading term.

Question

Differentiate with respect to x:

a) $(2x^2 + 5)^{{3 \over 2}}$

b) $\cos (5x^2 + 3)$

c) $(1 + 2\sin x)^4$

Answer

$\eqalign{ f(x) &=& (2x^2 + 5)^{{3 \over 2}} \cr \cr f'(x) &=& {3 \over 2}\;(2x^2 + 5)^{{1 \over 2}} \times 4x \cr \cr &=& 6x{\kern 1pt} \;(2x^2 + 5){\kern 1pt} ^{{1 \over 2}} \cr}$

$\eqalign{ f(x) &=& \cos \left( {5x^2 + 3} \right) \cr \cr f'(x) &=& - \sin \left( {5x^2 + 3} \right) \times 10x \cr \cr &=& - 10\;x\;\sin \left( {5x^2 + 3} \right) \cr}$

$\eqalign{ f(x) &=& (1 + 2\sin x)^4 \cr \cr f'(x) &=& 4\;(1 + 2\;\sin x)^3 (2\;\cos x) \cr \cr &=& 8\;\cos x\;(1\; + 2\sin x)^3 \cr}$

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Differentiation - test bite answers

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