Question

Question #a7bf9

Answer 1

`y'=[-3sin(3x)ln(x)+cos(3x)/x]x^cos(3x)`

Explanation:

Apply the natural log on both sides:

`y=x^(cos(3x))->lny=lnx^cos(3x)`

Since `ln(a^r)=rlna`, we can rewrite the problem as

`lny=cos(3x)*lnx`

Now we must use implicit differentiation `(dy/dx)` on both sides:

Differentiate both sides with respect to `x`

`1/y*dy/dx=color(red)(-3sin(3x)ln(x)+cos(3x)/x)larr`See proof below

`--------------------`
Apply the product rule:

`d/dx[f*g]=f'(x)*g(x)+g'(x)*f(x)`

Let: `f(x)=cos(3x)` and `g(x)=ln(x)`

Thus, `f'(x)=-3sin(3x)` and `g'(x)=1/x`

`d/dx[cos(3x)*ln(x)]=[-3sin(3x)*ln(x)]+([1/x]*[cos(3x)])`

`d/dx=-3sin(3x)ln(x)+cos(3x)/x`
`--------------------`

Going back to the problem:

`1/y*dy/dx=-3sin(3x)ln(x)+cos(3x)/x`

Multpliy `y` to both sides:

`cancelcolor(red)y*1/cancely*dy/dx=[-3sin(3x)ln(x)+cos(3x)/x]*color(red)y`

Since we want everything in terms of `x` we will replace `color(red)y` with the function first given since `color(red)y` is defined by the function `x^cos(3x)`. So finally,

`dy/dx=y'=[-3sin(3x)ln(x)+cos(3x)/x]x^cos(3x)`

Answer 2

Given: `y=x^cos(3x)`

Use the natural logarithm on both sides:

`ln(y) = ln(x^cos(3x))`

Use the property `ln(a^c) = (c)ln(a)`:

Given: `ln(y)=cos(3x)ln(x)`

Differentiate both sides with respect to x:

`(d(ln(y)))/dx=(d(cos(3x)ln(x)))/dx`

The left side requires the use of the chain rule:

`(d(ln(y)))/dydy/dx=(d(cos(3x)ln(x)))/dx`

`1/ydy/dx=(d(cos(3x)ln(x)))/dx`

The right side requires the use of the product rule:

`1/ydy/dx=(d(cos(3x)))/dxln(x)+ cos(3x)(d(ln(x)))/dx`

`1/ydy/dx=-3sin(3x)ln(x)+ cos(3x)/x`

Multiply both sides by y:

`dy/dx=(-3sin(3x)ln(x)+ cos(3x)/x)y`

Substitute `y = x^cos(3x)`:

`dy/dx=(-3sin(3x)ln(x)+ cos(3x)/x)x^cos(3x)`