Question

How to calculate the derivative?

`(Q1)` If `y = (x^2cos(x))/(sin(x))`, find `(dy)/(dx)`.

`(Q2)` If `y = e^(alpha x)`, show `(d^ny)/(dx^n) = alpha^(n)e^(alpha x)`

Answer 1

Q1: `dy/dx=2xcotx-x^2csc^2x`

Q2: See explanation below

Explanation:

Question 1:

`y=(x^2cosx)/sinx`

We can simplify the expression using the trigonometric identity

`cotx=cosx/sinx`

`rArry=x^2cotx`

Now simply apply the product rule.

`dy/dx=(d/dxx^2)cotx+x^2(d/dxcotx)`

`dy/dx=2xcotx-x^2csc^2x`

Question 2:

`y=e^(alphax)`

`dy/dx=e^(alphax)*alpha=alphae^(alphax)` (by the chain rule)

`(d^2y)/dx^2=d/dx(dy/dx)=d/dx(alphae^(alphax))=alphae^(alphax)*alpha=alpha^2e^(alphax)`

`vdots` Continuing this pattern to the `n`th derivative

`(d^ny)/dx^n=d/dx(d^(n-1)y)/dx^(n-1)=d/dx(alpha^(n-1)e^(alphax))=alpha^(n-1)e^(alphax)*alpha=alpha^(n)e^(alphax)`

Answer 2

Here's question `2`.

Let's create a table with the first few derivatives of `y = e^(alphax)`.

`y' = alphae^(alphax)`

`y'' = alpha^2e^(alphax)`

`y''' = alpha^3e^(alphax)`

`(d^n y)/(dx^n) = alpha^n e^(alpha x)`

As required.

Hopefully this helps!

Answer 3

Q1: `y'=(x^2cosxsinx-(2xcosx-x^2sinx)(cosx))/sin^2x`. Q2: See below.

Explanation:

Q1 requires application of both the product rule and quotient rule.

Recall that if `y=f(x)/g(x), dy/dx=(g(x)f'(x)-f(x)g'(x))/(g(x))^2`

For `y=(x^2cosx)/sinx,` `f(x)=x^2cosx, g(x)=sinx`.

Differentiate each of the above to have them on hand when applying the quotient rule.

`f'(x)=x^2d/dxcosx+cosxd/dxx^2` (product rule)

`f'(x)=-x^2sinx+2xcosx`

`g'(x)=cosx`

So, using the above formula for the quotient rule,

`y'=(x^2cosxsinx-(2xcosx-x^2sinx)(cosx))/sin^2x`

Simplify:

`y'=(x^2cosxsinx-2xcos^2x+x^2sinxcosx)/sin^2x`

`y'=(2x^2sinxcosx-2xcos^2x)/sin^2x`

For Q2, simply calculate the first few derivatives to recognize the pattern:

`y=e^(alphax)`
Applying the chain rule to differentiation:
`y'=e^(alphax)*d/dxalphax=alphae^(alphax)`
`y''=alphae^(alphax)*d/dx(alphax)=alpha^2e^(alphax)`
`y'''=e^(alphax)*d/dx(alphax)=alpha^3e^(alphax)`

So, the pattern is

`(d^ny)/(dx^n)=a^n e^(alphax)`

Answer 4

`(Q1)` `=> (dy)/(dx) = (2cot(x) - xcsc^2(x))x`

`(Q2)` `=> " See below"`

Explanation:

`(Q1)` `y = (x^2cos(x))/(sin(x)) = x^2cot(x)`

`(dy)/(dx) = (d(x^2))/dxcot(x) + x^2(d (cot(x)))/(dx)`

`(dy)/(dx) = 2xcot(x) + x^2(-csc^2(x))`

`=> (dy)/(dx) = (2cot(x) - xcsc^2(x))x`

`(Q2)` `y = e^(alpha x)`

`(d^ny)/(dx^n) = d^n/dx^n e^(alpha x)`

`(d^ny)/(dx^n) = e^(alpha x)d^n/(dx^n)alpha^nx^n`

`(d^ny)/(dx^n) = e^(alpha x) alpha^n d^n/(dx^n)x^n`

`(d^ny)/(dx^n) = e^(alpha x) alpha^n *1`

`=>(d^ny)/(dx^n) = alpha^n e^(alpha x)`

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For math formatting, see this

Answer 5

A more rigorous proof for `Q2` via Mathematical Induction:

Induction Proof - Hypothesis

We seek to prove that:

`(d^(n))/(dx^n) e^(alpha x)= alpha^n e^(alpha x)` ..... [A]

So let us test this assertion using Mathematical Induction:

Induction Proof - Base case:

We will show that the given result, [A], holds for `n=1`

When `n=1` we have, via the chain rule that:

`d/dx e^(alphax) = e^(alphax) d/dx (alphax)`
`\ \ \ \ \ \ \ \ \ \ = e^(alphax) d(alpha)`

Or more precisely:

`(d^(1))/(dx^1) e^(alphax) = d/dx = e^(alphax) = alpha^1e^(alphax)`

So the given result is true when `n=1`.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when `n=m`, for some `m in NN, m gt 1`, in which case for this particular value of `m` we have:

`(d^(m))/(dx^m) e^(alpha x)= alpha^m e^(alpha x)` ..... [B]

We can now differentiate [B] a further time, and again apply the chain rule, to get:

`d/dx ((d^(m))/(dx^m) e^(alpha x) ) = d/dx alpha^m e^(alpha x)`
`" " = alpha^m d/dx e^(alpha x)`
`" " = alpha^m \ e^(alpha x) \ d/dx (alpha x)`
`" " = alpha^m \ e^(alpha x) \ (alpha)`
`" " = alpha^(m+1) \ e^(alpha x)`

Which is the given result [A] with `n=m+1`

Induction Proof - Summary

So, we have shown that if the given result [A] is true for `n=m`, then it is also true for `n=m+1`, ie, that:

`(d^(m))/(dx^m) e^(alpha x) = alpha^m e^(alpha x) => (d^(m+1))/(dx^(m+1)) e^(alpha x) = alpha^(m+1) e^(alpha x)`

where `m gt 1`. But we initially showed that the given result was true for `n=1`, ie that:

`(d^(1))/(dx^1) e^(alpha x) = alpha^1 e^(alpha x)`

so it must also be true for `n=2, n=3, n=4, ...` and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for `n in NN`

Hence we have:

`(d^(n))/(dx^n) e^(alpha x)= alpha^n e^(alpha x) \ \ \ QED`