Urgent! Please help me! I have an exam in two days and I really need help with these questions. Please explain the steps and answers as thoroughly as possible.

Question

1. Find the derivative f'(x).

(a) f(x)=xe^xcosx

(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-3x²)^-5

(b) f(x)=x(square root of 1+x²)

(c) f(x)=cos²(x²)

 

5. Find the derivative f'(x).

(a) f(x)=lnx/x

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

(c) f(x)=x^lnx

Answer 1

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^xcosx)'=e^xcosx+xe^xcosx-xe^xsinx

 

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^-2ln(x)+x^-1*x^-1=-x^-2ln(x)+x^-2

 

Good luck on your exam! :)

Answer 2

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 (sec²x) [Product Rule

= sec x tan²x + sec³x

= sec x (tan²x + sec²x)

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 + cos²x + sin²x) / (1 + cos x)²

= (1 + cos x) / (1 + cos x)² [sin²x + cos²x = 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+x²)

f'(x) = 1√(1+x²) + x . x / √(1+x²) [Product & Chain Rules

= √(1+x²) + x² / √(1+x²)

= ((1+x²) + x² ) / √(1+x²)

= (2x² + 1) / √(1+x²)

4(c) f(x) = cos²(x²)

= cos²(u) with u(x) = x² => u'(x) = 2x

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

= -4x cos (x²) sin(x²) [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 + u²) [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}