Question

Compute f(x+h)-f(x)/h and then Lim h _ 0 f(x+h)-f(x)/h for the following questions F(x)=7x+12 f(x)=x^3
f(x)=5x^2-2x+4

Answer 1

$\large \bf { \underline{\underline{ {1}^{st} \: \: part : }}}$

Compute First law of derivative :

• Before getting into derivation let's understand what derivative actually mean. In a single word it's the slope. When secant line gradually approaches tangent line it's called derivative of that particular point.

Derivation :

Let, x is a point where we need to find the derivative of this function. In other word, we have to find out the slope on ( x , f( x ) ) in the graph.

We have taken a pt right from x at the distance h; ( x + h ).And AB is secant line which is gradually approaching tangent line. Means h approaches 0.

Slope will be {f( x + h ) - f( x )}/( x + h ) - x =>{f( x + h ) - f( x )}/h

Similarly when secant line gradually approaches tangent line from left it will be {f( x - h ) - f( x )}/(-h)

No doubt it's for secant line, but when we are taking about tangent line it will be

$\sf \lim_{∆x \to \: 0} \frac{∆f(x)}{∆x} \\ \\ = { \underline{ \boxed{\bf{\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} }}}}$

$\large \bf { \underline{\underline{ {2}^{nd} \: \: part : }}}$

$\: \bull \: \: \: \: \sf \: f( x ) = 7x+12$

$\mapsto \: \: \sf \: {f}^{'} ( x ) = \lim_{x\to 0 }\frac{ 7(x + h)+12 - (7x + 12)}{h}$

$\mapsto \: \: \sf \: {f}^{'} ( x ) = \lim_{x\to 0 }\frac{ \cancel{7x }+7 h + \cancel{12} - \cancel{ 7x } - \cancel{ 12}}{h}$

$\mapsto \: \: \sf \: {f}^{'} ( x ) = \lim_{x\to 0 }\frac{ 7 \cancel{h }}{ \cancel{h}}$

$\mapsto \: \: \bf \: {f}^{'} ( x ) = 7$

$\: \bull \: \: \: \: \sf \: g(x) = {x}^{3}$

$\mapsto \: \: \sf \: {g}^{'} (x) = \lim_{x\to 0 }\frac{{(x + h)}^{3} - {x}^{3} }{h}$

$\mapsto \: \: \sf \: {g}^{'} (x) = \lim_{x\to 0 }\frac{ {(x + h - x)}^{3} - 3x(x + h)({x} + h - {x}) }{h}$

$\mapsto \: \: \sf \: {g}^{'} (x) = \lim_{x\to 0 }\frac{ {( \cancel{x} + h - \cancel{ x})}^{3} - 3x(x + h) ( \cancel{x} + h - \cancel{x})}{h}$

$\mapsto \: \: \sf \: {g}^{'} (x) = \lim_{x\to 0 }\frac{ h\big \{ {h}^{2} - 3x(x + h) \big \}}{h}$

$\mapsto \: \: \bf \: {g}^{'} (x) = - 3 {x}^{2}$

$\: \bull \: \: \: \: \sf \: z(x)=5x^2-2x+4$

$\mapsto \sf \: \: {z}^{'} (x) = \lim_{x\to 0 } \frac{5{(x + h) }^{2} - 2(x + h) + 4 - \{5 {x}^{2} - 2x + 4 \} }{h}$

$\mapsto \sf \: \: {z}^{'} (x) = \lim_{x\to 0 } \frac{5 {x}^{2} +2xh + {h}^{2} - 2x - 2h + 4 - 5 {x}^{2} + 2x - 4 }{h}$

$\mapsto \sf \: \: {z}^{'} (x) = \lim_{x\to 0 } \frac{ {h}^{2} + 2xh - 2h}{h}$

$\mapsto \sf \: \: {z}^{'} (x) = \lim_{x\to 0 }(h + 2x - 2)$

$\mapsto \bf \: \: {z}^{'} (x) = 2x - 2$