Download App

CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following functions from first principles:
$x^2e^x.$
✓

Answer

Let $f(x) = x^2e^x$
$\Rightarrow f(x + h) = (x + h)^2 e^{(x+h)}$
$\therefore \frac{\text{d}}{\text{dx}}\{\text{f(x)}\}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}+\text{h})^2\text{e}^{(\text{x}+\text{h})}-\text{x}^2\text{e}^{\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{x}^2\text{e}^{(\text{x}+\text{h})}-\text{x}^2\text{e}^\text{x}}{\text{h}}+\frac{2\text{xhe}^{(\text{x}+\text{h})}}{\text{h}}+\frac{\text{h}^2\text{e}^{(\text{x}+\text{h})}}{\text{h}}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\bigg(\frac{\text{x}^2\text{e}^\text{x}\big(\text{e}^{(\text{x}+\text{h}-\text{x})}-1\big)}{\text{x}}+2\text{xe}^{(\text{x}+\text{h})}+\text{he}^{(\text{x}+\text{h})}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{x}^2\text{e}^{\text{x}}\frac{\big(\text{e}^\text{h}-1\big)}{\text{h}}+2\text{xe}^{(\text{x}+\text{h})}+\text{h}^{\text{e}}(\text{x}+\text{h})\bigg]$
$=\text{x}^2\text{e}^\text{x}+2\text{xe}^\text{x}+0\times\text{e}^\text{x}\ \Big[\text{Since,}\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
So,
$\frac{\text{d}}{\text{dx}}(\text{x}^2\text{e}^\text{x})=\text{e}^\text{x}(\text{x}^2+2\text{x})$

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 "5 Marks Questions" from CONTINUITY AND DIFFERENTIABILITY→CONTINUITY AND DIFFERENTIABILITY — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

$\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$, find AB. Hence, solve the system of equations:
x - 2y = 10, 2x + y + 3z = 8 and -2y + z = 7
→
Find the vector and Cartesian equations of the line through the point (5, 2, -4) and which is parallel to the vector $3\hat i + 2\hat j - 8\hat k$.
→
Find all points of discontinuity of f, where
$\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},\text{if x}<0\\ \text{x}+ 1, \text{if} \text{x}\geq0\end{cases}$
→
Write a value of $\int\text{e}^{\log\sin\text{x}}\cos\text{x}\text{ dx}$
→
Find the particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}+2\text{y}\ \tan\text{x}=\sin\text{x},$ given that $\text{y}=0\ \text{when x}=\frac{\pi}{3}.$
→
Find the area bounded by the curve $\text{y}=\cos\text{x}$, x-axis and the ordinates $\text{x}=0, \text{x}=2\pi$.
→
Solve the following equation for x:
$\tan^{-1}\frac{1}{4}+2\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{6}+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{4}$
→
If $\begin{vmatrix}\text{a}&\text{b}-\text{y}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0,$ then using properties of determinants, find the value of $\frac{\text{a}}{\text{x}}+\frac{\text{b}}{\text{y}}+\frac{\text{c}}{\text{z}},$ where $\text{x},\text{y},\text{z}\neq0.$ 
→
Prove that $\frac{1}{2} \tan ^{-1} x=\cos ^{-1}\left\{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}\right\}^{\frac{1}{2}}$
→
Evaluate the following integrals:
$\int\text{x}\sqrt{\text{x}^2+\text{x}}\text{dx}$
→