Differentiate the following from first principles:
(a) #y=10x#
(b) #y=6-5x+x^2#?

Question

Emilia Aguilar · Accepted Answer

a) #dy/dx = 10#

b) #dy/dx= -5+2x # By definition of the derivative: # dy/dx = f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h # Part (A) With # y=10x # we have; # dy/dx = lim_(h rarr 0) ( {10(x+h)} -{10x} ) / h # # " " = lim_(h rarr 0) ( 10x+10h-10x ) / h # # " " = lim_(h rarr 0) ( 10h ) / h # # " " = lim_(h rarr 0) ( 10 )# # " " = 10# Part (B) With # y=6-5x+x^2 # we have; # dy/dx = lim_(h rarr 0) ( {6-5(x+h)+(x+h)^2} -{6-5x+x^2} ) / h # # " " = lim_(h rarr 0) ( 6-5x-5h+x^2+2hx+h^2 -6+5x-x^2 ) / h # # " " = lim_(h rarr 0) ( -5h+2hx+h^2 ) / h # # " " = lim_(h rarr 0) ( -5+2x+h ) # # " " = -5+2x #

Ezekiel Alexander · Answer

undefined (a) The derivative of y = 10x from first principles is:

dy/dx = lim(h -> 0) [(10(x + h) - 10x) / h]
     = lim(h -> 0) [(10x + 10h - 10x) / h]
     = lim(h -> 0) [10h / h]
     = lim(h -> 0) 10
     = 10

(b) The derivative of y = 6 - 5x + x^2 from first principles is:

dy/dx = lim(h -> 0) [(6 - 5(x + h) + (x + h)^2 - (6 - 5x + x^2)) / h]
     = lim(h -> 0) [(6 - 5x - 5h + x^2 + 2xh + h^2 - 6 + 5x - x^2) / h]
     = lim(h -> 0) [(5x^2 - 5h + 2xh + h^2) / h]
     = lim(h -> 0) [5x^2/h - 5 + 2x + h]
     = 2x - 5

Samantha Adkison · Answer

undefined (a) To differentiate $ y = 10x $ from first principles:

Using the definition of the derivative from first principles:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Substitute the function $ f(x) = 10x $:

$$ f'(x) = \lim_{h \to 0} \frac{10(x + h) - 10x}{h} $$

Simplify:

$$ f'(x) = \lim_{h \to 0} \frac{10x + 10h - 10x}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{10h}{h} $$

Cancel out $ h $:

$$ f'(x) = \lim_{h \to 0} 10 $$

Evaluate the limit:

$$ f'(x) = 10 $$

Therefore, the derivative of $ y = 10x $ with respect to $ x $ from first principles is $ 10 $.

(b) To differentiate $ y = 6 - 5x + x^2 $ from first principles:

Using the definition of the derivative from first principles:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Substitute the function $ f(x) = 6 - 5x + x^2 $:

$$ f'(x) = \lim_{h \to 0} \frac{6 - 5(x + h) + (x + h)^2 - (6 - 5x + x^2)}{h} $$

Expand and simplify:

$$ f'(x) = \lim_{h \to 0} \frac{6 - 5x - 5h + x^2 + 2xh + h^2 - 6 + 5x - x^2}{h} $$
$$ f'(x) = \lim_{h \to 0} \frac{-5h + 2xh + h^2}{h} $$

Factor out $ h $:

$$ f'(x) = \lim_{h \to 0} \frac{h(-5 + 2x + h)}{h} $$

Cancel out $ h $:

$$ f'(x) = \lim_{h \to 0} (-5 + 2x + h) $$

Evaluate the limit:

$$ f'(x) = -5 + 2x $$

Therefore, the derivative of $ y = 6 - 5x + x^2 $ with respect to $ x $ from first principles is $ -5 + 2x $.

Differentiate the following from first principles: (a) #y=10x# (b) #y=6-5x+x^2#?