Find the derivative of px -q rx -s where p- q- r and s are fixed non-zero constants-

Let f (x) = nbsp;(px +q) ( rx +s ) ⇒ f(x) =pr +psx + qrx +qs ⇒ f(x) =psx + qrx +qs +pr Differentiating with respect to x ⇒ f'(x)=ps qrx^2+0 ∴ f'(x) =ps ...

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Standard XII

Physics

Basic Differentiation Rule

Find the deri...

Question

Find the derivative of $(p x + q) (\frac{r}{x} + s)$ where $p, q, r$ and $s$ are fixed non-zero constants.

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Solution

Let $f (x) = (p x + q) (\frac{r}{x} + s)$
$\Rightarrow f (x) = p r + p s x + \frac{q r}{x} + q s$
$\Rightarrow f (x) = p s x + \frac{q r}{x} + q s + p r$
Differentiating with respect to $x$
$\Rightarrow f^{'} (x) = p s - \frac{q r}{x^{2}} + 0$
$∴ f^{'} (x) = p s - \frac{r q}{x^{2}}$

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Basic Differentiation Rule

Standard XII Physics

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Find the derivative of (px+q)(rx+s) where p,q,r and s are fixed non-zero constants.

Let f(x)=(px+q)(rx+s) ⇒f(x)=pr+psx+qrx+qs ⇒f(x)=psx+qrx+qs+pr Differentiating with respect to x ⇒f′(x)=ps−qrx2+0 ∴f′(x)=ps−rqx2

Find the derivative of $(p x + q) (\frac{r}{x} + s)$ where $p, q, r$ and $s$ are fixed non-zero constants.

Let $f (x) = (p x + q) (\frac{r}{x} + s)$
$\Rightarrow f (x) = p r + p s x + \frac{q r}{x} + q s$
$\Rightarrow f (x) = p s x + \frac{q r}{x} + q s + p r$
Differentiating with respect to $x$
$\Rightarrow f^{'} (x) = p s - \frac{q r}{x^{2}} + 0$
$∴ f^{'} (x) = p s - \frac{r q}{x^{2}}$