Physics, asked by tanishq7940, 1 year ago

4.
ex sin x log x

differentiate using product rule​

Answers

Answered by PSN03
19

y=e^x sinx logx

dy/dx=sinx logx de^x/dx + e^x logx dsinx/dx + e^x sinx dlogx/dx

=sinx logx e^x + e^x logx cosx + e^x sinx 1/x

(as differentiation of sinx is cosx

differentiation of logx is 1/x

and differentiation of e^x is e^x)

Answered by swethassynergy
3

The  derivative  of the function e^{x}\ sinx\ logx is e^{x}  [\ \frac{six}{x} + \ cosx\ logx+ sinx\ logx].

Explanation:

Given:

The function e^{x}\ sinx\ logx.

To Find:

The  derivative  of the function e^{x}\ sinx\ logx.

Formula used:

A function y = u v, where u and v are the functions of x. Then, by the use of the product rule, we can easily find out the derivative of y with respect to x, and can be written as:

(dy/dx) = u (dv/dx) + v (du/dx)

Solution:

As given,the function e^{x}\ sinx\ logx.

y=e^{x}\ sinx\ logx

\frac{dy}{dx} =\frac{d}{dx}(e^{x})(\ sinx\ logx )

     = e^{x} \frac{d}{dx}(\ sinx\ logx )+sinx\ logx\times\frac{d}{dx} e^{x}

     = e^{x} [\frac{d}{dx}(\ sinx\ logx )]+sinx\ logx\times e^{x}

     =e^{x} [sinx\frac{d}{dx}(\ logx )+logx\frac{d}{dx}(\ sinx )]+sinx\ logx\times e^{x}

     =e^{x} [sinx \times \frac{1}{x} +logx\times cosx]+sinx\ logx\times e^{x}

     =e^{x}  \ \frac{six}{x} +e^{x} \ cosx\ logx+e^{x} sinx\ logx

      =e^{x}  [\ \frac{six}{x} + \ cosx\ logx+ sinx\ logx]

Thus,the  derivative  of the function e^{x}\ sinx\ logx is e^{x}  [\ \frac{six}{x} + \ cosx\ logx+ sinx\ logx].

#SPJ2

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