d dx [ e^ ax (bx + c)]= - Vidyadip

MCQ

${d \over {dx}}[{e^{ax}}\cos (bx + c)]=$

✓
${e^{ax}}[a\cos (bx + c) - b\sin (bx + c)]$
B
${e^{ax}}[a\sin (bx + c) - b\cos (bx + c)]$
C
${e^{ax}}[\cos (bx + c) - \sin (bx + c)]$
D
None of these

✓

Answer

Correct option: A.

${e^{ax}}[a\cos (bx + c) - b\sin (bx + c)]$

a
(a) $\frac{d}{{dx}}[{e^{ax}}\cos (bx + c)]$=$\frac{{dx}}{{dt}} = - 2\sin t + 2\sin 2t$

=${e^{ax}}[a\cos (bx + c) - b\sin (bx + c)]$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 5. continuity and differentiation→STD 12 - 5. continuity and differentiation — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

The value of $\cot \left(\sum_{n=1}^{23} \cot ^{-1}\left(1+\sum_{k=1}^n 2 k\right)\right)$ is

If $y = {e^{(1 + {{\log }_e}x)}}$, then the value of ${{dy} \over {dx}} = $

Let $f(x) = \left\{ \begin{array}{l}\frac{{x - 4}}{{|x - 4|}} + a,\;x < 4\\\,\,\,\,\,\,\,\,\,\,\,\,a + b,\,x = 4\\\frac{{x - 4}}{{|x - 4|}} + b,\,x > 4\end{array} \right.$. Then $f(x)$ is continuous at $x = 4$ when

Consider the curves $\text{y}=\sin\text{x}$ and $\text{y}=\cos\text{x}.$ What is the area of the region bounded by the above two curves and the lines $\text{x}=0$ and $\text{x}=\frac{\pi}{4}?$

Find the maximum value of $f(x)=\sin (\sin x)$ for all $x \in R$.

A box $B_1$ contains $1$ white ball, $3$ red balls and $2$ black balls. Another box $B_2$ contains $2$ white balls, $3$ red balls and $4$ black balls. A third bo $B _2$ contains $3$ white balls, $4$ red balls and $5$ black balls.

$1.$ If $1$ ball is drawn from each of the boxes $B_1, B_2$ and $B_3$, the probability that all $3$ drawn balls are of the same colour is

$(A)$ $\frac{82}{648}$ $(B)$ $\frac{90}{648}$ $(C)$ $\frac{558}{648}$ $(D)$ $\frac{566}{648}$

$2.$ If $2$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these $2$ balls are drawn from bo $B _2$ is

$(A)$ $\frac{116}{181}$ $(B)$ $\frac{126}{181}$ $(C)$ $\frac{65}{181}$ $(D)$ $\frac{55}{181}$

Give the answer question $1$ and $2.$

If $R(t) = \left[ {\begin{array}{*{20}{c}}{\cos t}&{\sin t}\\{ - \sin t}&{\cos t}\end{array}} \right],$then $R(s).\,R(t) = $

If $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x }>\frac{\pi}{2}\end{cases}$ Then f(x) is continuous at $\text{x}=\frac{\pi}{2},$ if:

For $x > 1$, if ${\left( {2x} \right)^{2y}} = 4{e^{2x - 2y}}$, then ${\left( {1 + {{\log }_e}\,2x} \right)^2}\frac{{dy}}{{dx}}$ is equal to

Which of the following differentials equation has $\text{y}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}$ as the general solution?