- #1
FaraDazed
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Homework Statement
Differentiate the following and display in the simplest form.
Part A:
[tex]y=0.2x^5 - sin4x + cos4x[/tex]
Part B:
[tex]y=(2x^4 +3)^3
[/tex]
Part C:
[tex]
y=2x^3 sinx
[/tex]
Part D:
[tex]
y=\frac{sinx}{x^2}
[/tex]
Part E:
[tex]
y=\frac{x^3}{x^2 +1}
[/tex]
Homework Equations
chain rule
product rule
quotient rule
The Attempt at a Solution
Part A:
[tex]
y=0.2x^5 - sin4x + cos4x \\
y'=x^4-4 \ cos4x-4 \ sin4x
[/tex]
Part B:
[tex]
y=(2x^4 +3)^3 \\
let \ u =2x^4 +3 \\
u'=8x^3 \\
y'(u)=3u^2 \\
y'=3u^2 \times 8x^3 \\
y'=3(2x^4 +3)^2 \times 8x^3 \\
y'=24x^3(2x^4+3)^2 \\
[/tex]
Part C:
[tex]
y=2x^3 \ sinx \\
let \ u = 2x^3 \\
let \ v = sinx \\
y'=vu'+uv' \\
y'=sinx \ 6x^2 + 2x^3 \ cosx \\
y'=2x^2(3 \ sinx + x \ cosx)
[/tex]
Part D:
[tex]
y=\frac{sinx}{x^2} \\
y'=\frac{vu'-uv'}{v^2} \\
y'=\frac{x^2 cosx - sinx \ 2x}{(x^2)^2} \\
y'=\frac{x(x \ cosx - 2sinx)}{x^4} \\
y'=x^{-3}(x \ cosx - 2sinx)
[/tex]
Part E:
[tex]
y=\frac{x^3}{x^2 +1} \\
y'=\frac{vu'-uv'}{v^2} \\
y'=\frac{(x^2 +1)3x^2-x^32x}{(x^2 + 1)^2} \\
y'=\frac{3x^4 + 3x^2 - 2x^4}{(x^2 + 1)^2} \\
y'=\frac{x^2(3x^2 + 3 - 2x^2 )}{(x^2 +1)^2} \\
[/tex]
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