Andre W. answered • 10/16/13

Friendly tutor for ALL math and physics courses

^{x}cosx)'=e

^{x}cosx+xe

^{x}cosx-xe

^{x}sinx

^{-5}

^{-6}*(-2-6x)

^{-1}*ln(x))' =-x

^{-2}ln(x)+x

^{-1}*x

^{-1}=-x

^{-2}ln(x)+x

^{-2}

Leigh A.

asked • 10/16/131. Find the derivative f'(x).

(a) f(x)=xe^{x}cosx

(b) f(x)=secxtanx

(c) f(x)=sinx/1+cosx

2. Let g(x) be a differentiable function such that g(0)=2 and g'(0)=3.

(a) f(x)=(g(x))^{3}

(b) f(x)=g(7x)

3. If h(t)=300+12sin(t/2), find h"(pi/3).

4. Find the derivative f'(x).

(a) f(x)=(4-2x-3x^{2})^{-5}

(b) f(x)=x(square root of 1+x^{2})

(c) f(x)=cos^{2}(x^{2})

5. Find the derivative f'(x).

(a) f(x)=lnx/x

(b) f(x)=arctan(e^{x})

(c) f(x)=x^{lnx}

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Andre W. answered • 10/16/13

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5
(3)
Friendly tutor for ALL math and physics courses

I'll do part (a) of each problem and give somebody else the chance to do the rest. :)

1a. Use the product rule, (uv)'=u'v+uv'.

For a product of 3 functions, (uvw)'=u'vw+uv'w+uvw'

(xe^{x}cosx)'=e^{x}cosx+xe^{x}cosx-xe^{x}sinx

2a. Use the chain rule, f(g(x))'=f'(g(x)) g'(x).

(g(x)³)'=3g(x)² g'(x)

(g³)'(0)=3g(0)² g'(0)=3*2²*3=36

3. h(t)=300+12sin(t/2)

h'(t)=6cos(t/2)

h''(t)=-3sin(t/2)

h''(Pi/3)=-3sin(Pi/6)=-3/2

4a. Chain rule again: outside derivative * inside derivative

f(x)=(4-2x-3x²)^{-5}

f'(x)=-5(4-2x-3x²)^{-6}*(-2-6x)

5a. Quotient rule or product rule, I'll use the product rule:

(ln(x)/x)'=(x^{-1}*ln(x))' =-x^{-2}ln(x)+x^{-1}*x^{-1}=-x^{-2}ln(x)+x^{-2}

Good luck on your exam! :)

Jeff K. answered • 05/22/20

Tutor

New to Wyzant
Together, we build an iron base of understanding

Hi Andre:

Here is how to solve parts (b) and (c)

1(b) f(x) = sec x tan x

f'(x) = (sec x tan x)tan x + sec x (sec^{2}x) [Product Rule

= sec x tan^{2}x + sec^{3}x

= sec x (tan^{2}x + sec^{2}x)

1(c ) f(x) = sin x / (1+cos x)

f'(x) = [(1 + cos x) cos x - sin x (-sin x) ] / (1 + cos x)^{2 } [Quotient Rule

= (cos x + cos^{2}x + sin^{2}x) / (1 + cos x)^{2}

^{ } = (1 + cos x) / (1 + cos x)^{2} [sin^{2}x + cos^{2}x = 1, trig identity

= 1 / (1 + cos x) [ Cancelling (1 + cos x)

2(b) f(x) = g(7x)

g'(7x) = 7 g'(x)

f'(x) = f'(g(x)) g'(x)

f'(0) = f'(g(0)) g'(0)

= f'(2). 3

= 3 f'(2)

4(b) f(x) = x√(1+x^{2})

f'(x) = 1√(1+x^{2}) + x . x / √(1+x^{2}) [Product & Chain Rules

= √(1+x^{2}) + x^{2} / √(1+x^{2})

= ((1+x^{2}) + x^{2 }) / √(1+x^{2})

= (2x^{2} + 1) / √(1+x^{2})

4(c) f(x) = cos^{2}(x^{2})

= cos^{2}(u) with u(x) = x^{2} => u'(x) = 2x

f'(x) = 2 cos (u) (-sin (u) . 2x [Chain Rule

= -4x cos (x^{2}) sin(x^{2}) [Substituting back for u = u(x)

5(b) f(x) = arctan(e^{x})

f(u) = arctan u putting u(x) = e^{x} => u' = e^{x}

f'(u) = (1 / (1 + u^{2}) [Derivative of arctan

f'(x) = f'(u) . u'(x) [Chain Rule^{ }

= 1 / (1 + (e^{x})^{2}) . e^{x}

= e^{x} / (1 + e^{2x})

5(c) f(x) = x^{ln x}

Let f(x) = y = x^{ln x} [when an exponent (power) includes a function of x, __use logs__

=> log y = (ln x) (ln x) [ By the laws of logs

= (ln x)^{2}

=> (1/y) y' = 2 ln x . 1/x [Implicit differentiation on both sides with the Chain Rule

=> y' = 2y ln x / x

= 2x^{ln x} ln x / x [Substituting back for y = x^{ln x}

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