Question

Find dy/dx, if y = (sin x) ^x +(cos x) ^tan x​

Answer 1

Answer:

$\bold{\purple{\fbox{\red{Answer}}}}$

$y = {sinx}^{x} + {cosx}^{tanx}.........(1)$

$\: \: \: to \: find = \frac{dy}{dx}$

$y = {sinx}^{x} + {cosx}^{tanx}$

$by \: getting \: log \: both \: side$

$logy = log {sin}^{x} + log {cosx}^{tanx}$

$logy = xlogsinx + tanxlogcosx$

$by \: product\: rule$

$\frac{1}{y} \times \frac{dy}{dx} = logsinx \times \frac{1}{sinx} \times cosx + {secx}^{2} \times logcosx \: \times \frac{1}{cosx} \times - sinx$

$just \: simplify$

$\frac{dy}{dx} = y(logsinx \times \frac{cosx}{sinx} + {secx}^{2} \times logcosx \times - \frac{sinx}{cosx}$

$\frac{dy}{dx } = y(logsinx \times cotx + {secx}^{2} \times logcosx - tanx)$

$\frac{dy}{dx} = y(logsinx \: cotx - {secx}^{2}logcosx \: tanx)$

$from \: equation \: (1) \: answer \: will \: be$

$\frac{dy}{dx} = {sin}^{x} + {cosx}^{tanx} (logsinxcotx - {secx}^{2}logcosxtanx)$

☞ correct$\large\blue{\rm\:Answer}$

$\frac{dy}{dx } = {sin}^{x} + {cosx}^{tanx} (logsinxcotx - {secx}^{2} logcosxtanx)$