Search any question & find its solution
Question:
Answered & Verified by Expert
If $y=\frac{e^{\sin x}+\sin \mathrm{h}^3 x}{\cosh x-\tan x}$, then $\mathrm{y}^{\prime}(0)=$
Options:
Solution:
2991 Upvotes
Verified Answer
The correct answer is:
2
Given $y=\frac{e^{\sin x}+\sinh ^3 x}{\cos n x-\tan x}$
Now, Differentiate w.r.t ' $x$ ' both sides, $\left(e^{\sin x} \cos x+3(\sinh x)^2 \cos x\right)$
$(\cosh x-\tan x)$
$-\left(e^{\sin x}+\sinh ^3 x\right)$
$\frac{\left(\sinh x+\sec ^2 x\right)}{(\cosh x-\tan x)^2}$
put $x=0$ in above derivative,
$\mathrm{y}^{\prime}(0)=\frac{\left[\begin{array}{l}\left(\mathrm{e}^{\sin 0}(\cos 0)+3(\sin 0)^2 \cos (0)\right)(\cos 0-\tan 0)- \\ \left(\mathrm{e}^{\sin 0}+(\sin (0))^3\left(\sin (0)-(\sec (0))^2\right)\right.\end{array}\right]}{(\cos (0)-\tan (0))^2}$
$\mathrm{y}^{\prime}(0)=\frac{[(1+0)(1-0)-(1+0)(0-1)]}{(1-0)^2}=\frac{1+1}{1}=2$
so,option (d) is correct.
Now, Differentiate w.r.t ' $x$ ' both sides, $\left(e^{\sin x} \cos x+3(\sinh x)^2 \cos x\right)$
$(\cosh x-\tan x)$
$-\left(e^{\sin x}+\sinh ^3 x\right)$
$\frac{\left(\sinh x+\sec ^2 x\right)}{(\cosh x-\tan x)^2}$
put $x=0$ in above derivative,
$\mathrm{y}^{\prime}(0)=\frac{\left[\begin{array}{l}\left(\mathrm{e}^{\sin 0}(\cos 0)+3(\sin 0)^2 \cos (0)\right)(\cos 0-\tan 0)- \\ \left(\mathrm{e}^{\sin 0}+(\sin (0))^3\left(\sin (0)-(\sec (0))^2\right)\right.\end{array}\right]}{(\cos (0)-\tan (0))^2}$
$\mathrm{y}^{\prime}(0)=\frac{[(1+0)(1-0)-(1+0)(0-1)]}{(1-0)^2}=\frac{1+1}{1}=2$
so,option (d) is correct.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.