Question

differentiate using first principle- (x+1)(2x-3)

Answer 1

There are two was to differentiate this:
The first would be performing the distributive property and derive term by term:
\[(x+1)(2x-3) \iff 2x^2-3x+2x-3\]

The second involves the chain rule or also called the derivative of a product of functions:
\[h(x)=f(x).u(x) \iff h'(x)=f'(x).u(x)+f(x).u'(x)\]

Answer 2

"differentiate using first principle" usually refers to finding the derivative using the definition.

Answer 3

That's new to me. Please describe the method you want or have to use as "differentiate using the definition of the derivative." If you type merely "differentiate," that means "differentiate, using the approrpriate differentiation formualas. "

I understand that there may be cultural differences here, and assure you I do not mean any offense.

Answer 4

@Zarkon, I agree with him totally correct. As refereed in http://www.everythingmaths.co.za/maths/grade-12/06-differential-calculus/06-differential-calculus-02.cnxmlplus

Zarkon.medals++;

Answer 5

Ah well, then my suggestions are no good.

Answer 6

\[(x+1)(2x-3)=2x^2-3x+2x-3 \\ Let\ f(x)= 2x^2-3x+2x-3= 2x^2-x-3\\ \\ \implies f(x+\delta \ x)= 2(x+\delta)^2-(x+\delta)-3\\ \implies \ f(x+\delta)-f(x)= [2(x+\delta)^2-(x+\delta)-3]- \ [2x^2-x-3]\\= 2x^2+2\delta^2+4x \delta-x-\delta-2x^2+x=3\\=2\delta^2+4x\ \delta-\delta\\ now\ \frac{ f(x+\delta)-f(x) }{ \delta }=\frac{ 2\delta^2+4x\ \delta-\delta }{ \delta }\\ \lim_{\delta \rightarrow 0}\frac{ f(x+\delta)-f(x) }{ \delta }=\lim_{\delta \rightarrow 0}\frac{ 2\delta^2+4x\ \delta-\delta }{ \delta }\\implies \ \frac{ df(x) }{ dx }=4x-1\]

Answer 7

Thanxs everyone for taking the trouble to clear my doubt.It's appreciated.I get the point when u say,i need to b more specific in stating my question.Will keep that in mind,when i post my questions in future.