Question

The nth derivative of $h(x) = {e^{3x + 5}}{x^2}$ at x=0 is
A.${e^5}{3^{n - 2}}n\left( {n - 1} \right)$
B.${e^5}{3^{n + 2}}n\left( {n - 1} \right)$
C.${e^5}{3^n}n\left( {n - 1} \right)$
D.${e^5}{3^{n - 2}}n\left( {n + 1} \right)$

Answer 1

Hint: We will find 1st differentiation and their corresponding value at x = 0, then 2nd differentiation and their corresponding value at x = 0, and so on. We will see a pattern. According to that pattern we can write the nth derivative at x = 0

Complete step-by-step answer:
Given:
\[h(x) = {e^{3x + 5}}{x^2}\]
Differentiating h(x) both sides, we get
\[h\prime (x) = 3{e^{3x + 5}}{x^2} + 2x{e^{3x + 5}}\]
\[h\prime \left( 0 \right) = 0\]
Again on differentiating h′(x), we get
\[h\prime \prime (x) = 3h\prime (x) + 6x{e^{3x + 5}} + 2{e^{3x + 5}}\]
\[h\prime \prime (x) = 3h\prime (x) + 6x{e^{3x + 5}} + 2{e^{3x + 5}}\]
\[h\prime \prime (x) = 3h\prime (x) + 3(h\prime (x) - h(x)) + 2{e^{3x + 5}}\]
\[h\prime \prime (x) = 6h\prime (x) - 3h(x) + 2{e^{3x + 5}}\]
\[h\prime \prime \left( 0 \right) = 2{e^5}\]
Differentiating h′′(x), we get
\[h\prime \prime \prime (x) = 6h\prime \prime (x) - 3h\prime (x) + 6{e^{3x + 5}}\]
\[h\prime \prime \prime \left( 0 \right) = 18{e^5}\]
... and so on.
According to the pattern, the value of nth derivative of h(x) at x=0 is
\[{h^n}\left( 0 \right) = {e^5}{3^{n - 2}}n(n - 1)\]
Hence, option A is the correct answer.

Additional information: For standard function,
Step :- Use simple differentiation to get 1st, 2nd and 3rd derivatives.
Step 2:- Observe the changes. Some of the changes are the power of the function, additional coefficients, increase in angle, etc.
Step 3:- Express it in it's Nth derivative form by the help of the changes observed. This will be your general formula for the Nth derivative of the standard function.
For non-standard functions,
Step 1:- Express them in one or more standard functions.
Step 2:-Use the general formula of Nth derivative of standard functions to find the Nth derivative of non-standard functions.

Note: Using the nth derivative result at x= 0 we can find the any derivative value by just simply putting the value of n on this final result. Like if we have to find the value of the 99th derivative value at x=0, then we will simply put the value of n=99 on the final result.