Differentiation Questions

Solution : Let y = $x^{sinx}$. Then, Taking log both sides, log y = sin x.log x $\implies$ y = $e^{sin x.log x}$ By using logarithmic differentiation, On differentiating both sides with respect to x, we get $dy\over dx$ = $e^{sin x.log x}$$d\over dx$(sin x.log x) $\implies$ $dy\over dx$ = \(x^{sin x}{log x {d\over dx}(sin […]

Solution : Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as y = $\sqrt{sin x + y}$ Squaring on both sides, $\implies$  $y^2$  = sin x + y By using differentiation of infinite series, Differentiating both sides with respect to x, 2y \(dy\over

Solution : We have, x = a{cos t + ${1\over 2} log tan^2 {t\over 2}$} and y = a sin t $\implies$ x = a{cos t + ${1\over 2} \times 2 log tan{t\over 2}$} and y = a sin t $\implies$ x = a{cos t + {$log tan{t\over 2}$} and y = a sin t

Solution : Let y = cosx.sinx By using product rule in differentiation, $dy\over dx$ = sinx(-sinx) + cosx.cosx $dy\over dx$ = $cos^2x – sin^2x$ = cos 2x Hence, the differentiation of cosx.sinx with respect to x is cos 2x. Questions for Practice What is the differentiation of $e^{sinx}$ ? What is the differentiation of sin

Solution : Let y = $e^{sinx}$. Putting u = sinx , we get y = $e^u$ and u = sinx $\therefore$  $dy\over du$ = $e^u$ and $du\over dx$ = cosx Now, $dy\over dx$ = $dy\over du$ $\times$ $du\over dx$ $\implies$ $dy\over dx$ = $e^u$cosx = $e^{sinx}$cosx Hence, the differentiation of $e^{sinx}$ with respect to x

Solution : The differentiation of sin square x with respect to x is sin 2x. Explanation : We have, y = $sin^2 x$ Differentiating by using chain rule, $dy\over dx$ = 2 sin x cos x $dy\over dx$ = sin 2x Hence, $dy\over dx$ = sin 2x Questions for Practice What is the differentiation of

Solution : We have, y = 1/sinx = cosecx By using differentiation formula of cosecx, $dy\over dx$ = -cosecx.cotx Hence, the differentiation of 1/sinx = -cosecx.cotx Questions for Practice What is the differentiation of $e^{sinx}$ ? What is the differentiation of sin square x or $sin^2x$ ? What is the differentiation of cosx sinx ?

Solution : We have, y = $sin x^2$ Differentiating with respect to x by using chain rule, $dy\over dx$ = $cos x^2$.(2x) $dy\over dx$ = 2x.$cos x^2$ Hence, the differentiation of $sin x^2$ with respect to x is 2x.$cos x^2$ Questions for Practice What is the differentiation of $e^{sinx}$ ? What is the differentiation of

Solution : We have, y = log sin x By using chain rule in differentiation, Let u = sin x $\implies$ $du\over dx$ = cos x And, y = log u $\implies$ $dy\over du$ = $1\over u$  Now, $dy\over dx$ = $dy\over du$ $\times$ $du\over dx$ $\implies$ $dy\over dx$ = $1\over u$ $\times$ cos x

Solution : We have, y = $1\over log x$ By using quotient rule in differentiation, $dy\over dx$ = $log x.{d\over dx}(1) – 1 {d\over dx}(log x)\over (log x)^2$ $dy\over dx$ = $0 – {1\over x}\over (log x)^2$ = $-1\over x (log x)^2$ Hence, the differentiation of 1/log x with respect to x is \(-1\over x