How to find formula for nth derivative of #f(x)=e^(2x)(x^2-3x+2)#?

3 Answers
Andrea S.
Jul 4, 2017

#d^n/dx^n (e^(2x)(x^2-3x+2)) = 2^n e^(2x)(x^2 +x(n-3)+(n^2-7n+8)/2)#

Explanation:

Pose:

#P(x) = x^2-3x+2#

The first derivative can be evaluated with the product rule:

#d/dx (e^(2x)(x^2-3x+2)) = e^(2x)(2x-3)+2e^(2x)(x^2-3x+2)#

#d/dx (e^(2x)(x^2-3x+2)) = e^(2x)(2x-3+2x^2-6x+4)#

#d/dx (e^(2x)(x^2-3x+2)) = e^(2x)(2x^2-4x+1)#

For #n>=2#, using the product rule for the #n#-th derivative:

#d^n/dx^n (e^(2x)P(x)) = sum_(k=0)^n ((n),(k)) d^k/dx^k (P(x))d^(n-k)/dx^(n-k) (e^(2x))#

Note now that, as #P(x)# is a polynomial of second degree, its derivative from the third order onward are identically null:

#d/dx P(x) = 2x-3#

#d^2/dx^2 P(x) = 2#

#d^k/dx^k P(x) = 0# for #k > 2#

So, for #n>=2# in the sum we can anyway ignore the terms for #k > 2#

#d^n/dx^n (e^(2x)P(x)) = sum_(k=0)^2 ((n),(k)) d^k/dx^k (P(x))d^(n-k)/dx^(n-k) (e^(2x))#

On the other hand:

#d^j/dx^j (e^(2x)) = 2^je^(2x)#

so:

#d^n/dx^n (e^(2x)P(x)) = sum_(k=0)^2 ((n),(k)) d^k/dx^k (P(x))2^(n-k)e^(2x)#

and as #e^(2x)# is common to all terms:

#d^n/dx^n (e^(2x)P(x)) = e^(2x)sum_(k=0)^2 2^(n-k) ((n),(k)) d^k/dx^k (P(x))#

There are only three terms in the sum and we can write them explicitly:

#k=0 => 2^(n-k) ((n),(k)) d^k/dx^k (P(x)) = 2^n((n),(0))P(x) = 2^n(x^2-3x+2)#

#k=1 => 2^(n-k) ((n),(k)) d^k/dx^k (P(x)) = 2^(n-1)((n),(1))(dP(x))/dx = 2^(n-1)n(2x-3) = 2^n(nx-(3n)/2)#

#k=2 => 2^(n-k)((n),(k)) d^k/dx^k (P(x)) = 2^(n-2)((n),(2))(d^2P(x))/dx^2 =2^(n-2)(n(n-1))/2 xx 2 = 2^n (n(n-1))/4#

Then:

#sum_(k=0)^2 ((n),(k)) d^k/dx^k (P(x))2^(n-k) = 2^n(x^2-3x+2+nx-(3n)/2+ (n(n-1))/4)#

simplifying:

#sum_(k=0)^2 ((n),(k)) d^k/dx^k (P(x))2^(n-k) = 2^n(x^2+x(n-3)+(8-6n+n^2-n)/4)#

#sum_(k=0)^2 ((n),(k)) d^k/dx^k (P(x))2^(n-k) = 2^n(x^2 +x(n-3)+(n^2-7n+8)/4)#

and in conclusion:

#d^n/dx^n (e^(2x)(x^2-3x+2)) = 2^n e^(2x)(x^2 +x(n-3)+(n^2-7n+8)/2)#

By direct inspection we can see the formula gives the right derivative also for #n=1# and for #n=0# (the function itself).

Steve M
Jul 4, 2017

# f^((n))(x) = {4x^2+4nx +n^2-7n -12x + 8} \ 2^(n-2) \ e^(2x) #

Explanation:

We have:

# f(x) = e^(2x)(x^2-3x+2) #

Although we could leave the function "as is" and use the product rule to differentiate, it makes more sense, to multiply out and analyze each component separately:

# f(x) = x^2e^(2x)-3xe^(2x)+2e^(2x) #
# " " = f_1(x) -3f_2(x) + 2f_3(x) #

say, where:

# f_1(x) = x^2e^(2x) #
# f_2(x) = xe^(2x) #
# f_3(x) = e^(2x) #

Then trivially, the #nth# derivative of #f(x)# is:

# f^((n))(x) = f_1^((n))(x) -3f_2^((n))(x) + 2f_3^((n))(x) # ..... [A]

#nth# derivative of #f_3(x)=e^(2x)#

Differentiating wrt #x# we have

# f_3^((1))(x) = 2e^(2x) \ \ \ \ \ \ \ \ \ \ = 2^1e^(2x)#
# f_3^((2))(x) = 2*2e^(2x) \ \ \ \ \ = 2^2e^(2x)#
# f_3^((3))(x) = 2*2*2e^(2x) = 2^3e^(2x)#

Leading us to conclude that:

# f_3^((n))(x) = 2^n \ e^(2x)# ..... [B]

#nth# derivative of #f_2(x) = xe^(2x)#

We have:

# f_2(x) = xf_3(x) #

Differentiating wrt #x#, and applying the product rule, we have

# f_2^((1))(x) = (x)(f_3^((1))(x)) + (1)(f_3(x)) #
# " " = x \ f_3^((1))(x) + f_3(x) #

# f_2^((2))(x) = (x)(f_3^((2))(x)) + (1)(f_3^((1))(x)) + f_3^((1))(x)#
# " " = x \ f_3^((2))(x) + 2 \ f_3^((1))(x)#

# f_2^((3))(x) = (x)(f_3^((3))(x)) + (1)(f_3^((2))(x)) + 2 \ f_3^((2))(x)#
# " " = xf_3^((3))(x) + 3 \ f_3^((2))(x)#

Leading us to conclude that:

# f_2^((n))(x) = xf_3^((n))(x) + n \ f_3^((n-1))(x)# ..... [C}

And using [B] this becomes:

# f_2^((n))(x) = x \ 2^n \ e^(2x) + n \ 2^(n-1) \ e^(2x)#
# " " = x \ 2 \ 2^(n-1) \ e^(2x) + n \ 2^(n-1) \ e^(2x)#
# " " = (2x+n) \ 2^(n-1) \ e^(2x) # ..... [D]

#nth# derivative of #f_1(x) = x^2e^(2x)#

We have:

# f_1(x) = xf_2(x) #

Using the above result [C] we conclude that:

# f_1^((n))(x) = xf_2^((n))(x) + n \ f_2^((n-1))(x)#

Using the result [D] this becomes:

# f_1^((n))(x) = x (2x+n) \ 2^(n-1) \ e^(2x) + n(2x+n-1) \ 2^(n-1-1) \ e^(2x) #
# " " = x (2x+n) \ 2^(n-1) \ e^(2x) + n(2x+n-1) \ 2^(n-2) \ e^(2x) #
# " " = 2x (2x+n) \ 2^(n-2) \ e^(2x) + n(2x+n-1) \ 2^(n-2) \ e^(2x) #
# " " = (2x (2x+n) + n(2x+n-1) ) \ 2^(n-2) \ e^(2x) #
# " " = (4x^2+2nx + 2nx+n^2-n ) \ 2^(n-2) \ e^(2x) #
# " " = (4x^2+4nx +n^2-n ) \ 2^(n-2) \ e^(2x) #

Combining the results:

Recall from [A] we have:

# f^((n))(x) = f_1^((n))(x) -3f_2^((n))(x) + 2f_3^((n))(x) #

Combining the above results we get:

# f^((n))(x) = (4x^2+4nx +n^2-n ) \ 2^(n-2) \ e^(2x) -3 (2x+n) \ 2^(n-1) \ e^(2x) + 2(2^n \ e^(2x)) #

# " " = (4x^2+4nx +n^2-n ) 2^(n-2) \ e^(2x) -3 (2x+n) \ 2 \ 2^(n-2) \ e^(2x) + 2 ( 2 \ 2 \ 2^(n-2) \ e^(2x)) #

# " " = {(4x^2+4nx +n^2-n ) -6 (2x+n) + 8} \ 2^(n-2) \ e^(2x)) } #

# " " = {4x^2+4nx +n^2-n -12x-6n + 8} \ 2^(n-2) \ e^(2x) #

# " " = {4x^2+4nx +n^2-7n -12x + 8} \ 2^(n-2) \ e^(2x) #

Ratnaker Mehta · Cesareo R.
Jul 5, 2017

# 2^(n-2)e^(2x){4x^2+4(n-3)x+(n^2-7n+8)}.#

Explanation:

We will solve this Problem, using the following Leibnitz Theorem

(LT) for #n^(th)# Derivative of Product of two functions #u and v.#

Denoting the #n^(th)# der. of fun. #u# by #u_n,# we have,

#(uv)_n=u_nv+""_nC_1u_(n-1)v_1+""_nC_2u_(n-2)v_2+...+u v_n.#

We take, #u=e^(2x) rArr u_n=2^n*e^(2x), and, v=x^2-3x+2.#

#:. v_1=2x-3, v_2=2, &, v_3=v_4=...=v_n=0, n>=3.#

#:. (d^n)/dx^n{(x^2-3x+2)e^(2x)}=(2^n*e^(2x))(x^2-3x+2)#
#+n*(2^(n-1)e^(2x))(2x-3)+n/2*(n-1)*(2^(n-2)e^(2x))*2,#

#=2^n*e^(2x)(x^2-3x+2)+2^(n-1)e^(2x)(2nx-3n)+2^(n-2)e^(2x)(n^2-n),#

#=2^(n-2)e^(2x){4(x^2-3x+2)+2(2nx-3n)+(n^2-n)},#

#=2^(n-2)e^(2x){4x^2+4(n-3)x+(n^2-7n+8)}.#

Enjoy Maths.!

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