$\begin{array}{1 1} (A)\;y'' = 4be^{2x} + 4e^{2x}(a+bx) \\ (B)\;y'' = 4be^{x} + 4e^{2x}(a+bx) \\ (B)\;y'' = 4be^{x} + 4e^{2x}(a+bx) \\(D)\;y'' = 4be^{2x} + 4e^{x}(a+bx)\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- When the function occurs in the form of their products, we can differentiate them using product rule, which states that $uv = uv' + vu' $

Step 1:

Given: $y = e^{2x}(a + bx)$

Differentiating on both sides by applying the product rule

let $u$ be $e^{2x}$ hence $u'$ will be $2e^2x$ and $v = a+bx$, hence $v' = b$

Hence $y' = 2e^{2x}(a+bx) + be^{2x}$

Step 2:

substituting $y = e^{2x}(ax+b)$

$y' = 2y + be^{2x}$

$y' - 2y = be^{2x}$-----(1)

Differentiating $y'$ again by applying the product rule we get

$y'' = 2be^{2x} + 4e^{2x}(a+bx) + 2e^{2x}(b)$

$y'' = 4be^{2x} + 4e^{2x}(a+bx)$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...