How to calculate the derivative?

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(Q1)(Q1) If y = (x^2cos(x))/(sin(x))y=x2cos(x)sin(x), find (dy)/(dx)dydx.

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

5 Answers
Mar 30, 2018

Q1: dy/dx=2xcotx-x^2csc^2xdydx=2xcotxx2csc2x

Q2: See explanation below

Explanation:

Question 1:

y=(x^2cosx)/sinxy=x2cosxsinx

We can simplify the expression using the trigonometric identity

cotx=cosx/sinxcotx=cosxsinx

rArry=x^2cotxy=x2cotx

Now simply apply the product rule.

dy/dx=(d/dxx^2)cotx+x^2(d/dxcotx)dydx=(ddxx2)cotx+x2(ddxcotx)

dy/dx=2xcotx-x^2csc^2xdydx=2xcotxx2csc2x

Question 2:

y=e^(alphax)y=eαx

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

(d^2y)/dx^2=d/dx(dy/dx)=d/dx(alphae^(alphax))=alphae^(alphax)*alpha=alpha^2e^(alphax)d2ydx2=ddx(dydx)=ddx(αeαx)=αeαxα=α2eαx

vdots Continuing this pattern to the nnth 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)dnydxn=ddxdn1ydxn1=ddx(αn1eαx)=αn1eαxα=αneαx

Mar 30, 2018

Here's question 22.

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

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!

Mar 30, 2018

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)

Mar 30, 2018

(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

Mar 30, 2018

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