Question

derivative of (sinx)logx+xsinx​

Answer 1

Answer:

sinx

------- + logx cosx +x cosx +sinx

x

Step-by-step explanation:

given--->

---------

(sinx) logx +x sinx

To find--->

------------

Derivative of given function

solution --->

--------------

formula used--->

--------------------

d

------ (sinx) =cosx

dx

d 1

-----(logx) =------

dx x

d

-----(x) = 1

dx

d dv du

------(u v) =u ------ + v -------

dx dx dx

now coming to question

let

y=(sinx) logx + x sinx

differentiating with respect to x

dy d d

-----=-------(sinx log x) +-----(x sinx)

dx dx dx

d d d

=sinx -----(logx) +logx----(sinx) +x----(sinx)

dx dx dx

d

+sinx ------(x)

dx

1

=sinx( -----)+logx cosx +x cosx +sinx (1)

x

dy sinx

----- =--------+logx cosx +x cox + sinx dx x x

Hope it helps you

Thanks

Answer 2

$\huge\underline\mathfrak\red{Answer}$

$( \sin \alpha ) log \alpha + \alpha \sin \alpha = y \\ \\ \frac{dy}{d \alpha } = \frac{d( \sin \alpha) }{d \alpha } log \alpha + \frac{d( log \alpha ) }{d \alpha } \sin \alpha + \alpha \frac{d (\sin \alpha )}{d \alpha } + \frac{d \alpha }{d \alpha } \sin \alpha \\ \\ \frac{dy }{d \alpha } = (\cos \alpha log \alpha + \frac{ \sin \alpha }{ \alpha } + \alpha \cos \alpha + \sin \alpha ) \frac{d \alpha }{d \alpha } \\ \\ dy = (\cos \alpha log \alpha + \frac{ \sin \alpha }{ \alpha } + \alpha \cos \alpha + \sin \alpha )d \alpha$