Differentiation (Part-38) | S N De

$~8.~~$ Differentiate the following functions w.r.t. t (or,$~u$) :

$~(i)~~(3^t\cos t+\sin t)$

Sol. Let $~~~~y=3^t\cos t+\sin t \\ \therefore \frac{dy}{dt}=\frac{d}{dt}(3^t)~\cos t+3^t~\frac{d}{dt}(\cos t)\\~~~~~~~~~~~~~+\frac{d}{dt}(\sin t)\\~~~~~~~~=(3^t ~\log_e 3)~\cos t+3^t(-\sin t)\\~~~~~~~~~~~~~~~~~~~~~~~~~+\cos t \\~~~~~~~~=3^t(\log_e3 ~\cos t-\sin t)+\cos t~~\text{(ans.)}$

$~(ii)~~(t^{-2}+t^2)(2^t-3^t)$

Sol. $~~y=(t^{-2}+t^2)(2^t-3^t) \\ \therefore~ \frac{dy}{dt}\\=\left[\frac{d}{dt}(t^{-2}+t^2)\right](2^t-3^t)\\~~~~~~~~~+(t^{-2}+t^2)\frac{d}{dt}(2^t-3^t)\\=[-2t^{-3}+2t](2^t-3^t)\\~~~~~~~~~+(t^{-2}+t^2)\left[\frac{d}{dt}(2^t)-\frac{d}{dt}(3^t)\right]\\=2(2^t-3^t)(t-t^{-3})\\~~~~~~~~~+(t^{-2}+t^2)(2^t\log_e2-3^t \log_e3)~~\text{(ans.)}$

$~~(iii)~~\sqrt te^t \sec t$

Sol. let $~~~~y=\sqrt te^t \sec t \\ \therefore~ \frac{dy}{dt}\\=\left[\frac{d}{dt}(\sqrt t)\right]e^t\sec t+\left[\frac{d}{dt}(e^t)\right]\sqrt t \sec t\\~~~~~~+\left[\frac{d}{dt}(\sec t)\right]\sqrt t e^t\\=\frac 12t^{-1/2}e^t\sec t+e^t\sqrt t\sec t\\~~~~~~+\sec t\tan t \sqrt t e^t\\=\frac{e^t\sec t}{2\sqrt t}(1+2t+2t \tan t)~~\text{(ans.)}$

$~(iv)~~\frac{u}{e^u-1}$

Sol. $~~y=\frac{u}{e^u-1} \\ \therefore \frac{dy}{du}\\=\frac{d}{du}\left(\frac{u}{e^u-1}\right)\\=\frac{(e^u-1)\frac{d}{du}(u)-u\frac{d}{du}(e^u-1)}{(e^u-1)^2}\\=\frac{(e^u-1) \cdot 1-u\cdot e^u}{(e^u-1)^2}\\=\frac{e^u(1-u)-1}{(e^u-1)^2}~~\text{(ans.)}$

To download full PDF containing full solution of DIFFERENTIATION (PART-1) [CLASS-XII] of Exercise-3A, click here.

$~(v)~~\frac{u\sin u+\cos u}{u^2 \log u}$

Sol. let $~~y=\frac{u\sin u+\cos u}{u^2 \log u}\\ \therefore \frac{dy}{du}\\=\frac{d}{du}\left(\frac{u\sin u+\cos u}{u^2 \log u}\right)\\=\frac{(u^2\log u)\frac{d}{du}(u\sin u+\cos u)\\~~~~~~-(u\sin u+\cos u)\frac{d}{du}(u^2\log u)}{(u^2\log u)^2}\\=\frac{u^2\log u (1 \cdot \sin u+u\cos u-\sin u)\\~~~~~-(u\sin u+\cos u)(u^2 \cdot \frac 1u+2u \log u)}{u^4(\log u)^2}\\=\frac{u^2\log u(\sin u+u\cos u-\sin u)\\~~~~~-(u\sin u+\cos u)(u+2u\log u)}{u^4(\log u)^2}\\=u \cdot \frac{u\log u(u\cos u)\\~~~~-(u\sin u+\cos u)(1+2\log u)}{u^4(\log u)^2}\\=\frac{u^2\log u\cos u\\~~~~~-(u\sin u+\cos u)(1+2\log u)}{u^3(\log u)^2}~~\text{(ans.)}$