Question
Question: If \(y = 1 - x + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} - .....\), th...
If y=1−x+2!x2−3!x3+4!x4−....., then dx2d2y is equal to
\eqalign{
& 1) - x \cr
& 2)x \cr
& 3)y \cr
& 4) - y \cr}
Solution
Hint : In the given question, we need to find dx2d2y, which is double differentiation or differentiating twice. There are factorials of 2 and 3 in the question, we do not have to find the values. While differentiating once, we only simplify such terms. Then we differentiate it again.
The formulas used to solve the problem are:
\eqalign{
& \dfrac{d}{{dx}}const = 0 \cr
& \dfrac{d}{{dx}}\left( x \right) = 1 \cr
& \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{\left( {n - 1} \right)}} \cr}
Complete step-by-step answer :
The given expression is as follows,
y=1−x+2!x2−3!x3+4!x4−.....
We do not know the last term, so we write is as it is given
Now, differentiating with respect x, we get
dxdy=−1+2!2x−3!3x2+4!4x3−.....
This can be done using the above-mentioned formulas.
Now, simplifying it further,
dxdy=−1+x−2!x2+3!x3−....
Which will be equal do
dxdy=−y
Now, we need to differentiate the above given expression to get dx2d2y
⇒dx2d2y=dx−dy
\eqalign{
& \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - ( - y) \cr
& \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = y \cr}
Therefore, the final answer is y.
Hence, option (3) is the correct answer.
So, the correct answer is “Option 3”.
Note : The question asks for double differentiation, even though we get a simple answer after differentiating once, do not forget to repeat the process. After the first differentiation, the answer is negative, so carry the sign as is it for the next step. Be careful about the signs and their derivatives. Learn the formula and apply them appropriately.