Differentiate from first principles if y=5
​

y'(x) = 7 ;

y'(x) = 15x²

Explanation:

Differentiation from first principle :

y = 7x

Formula to obtain derivative from first principle:

y'(x) = lim(h - - > 0) : [y(x + h) - y(x)] / h

Obtain : y(x + h)

y(x) = 7x

y(x + h) = 7(x + h)

Substitute into formula :

y'(x) = lim(h - - > 0) : [7(x + h) - 7x] / h

Expand :

y'(x) = lim(h - - > 0) : [7x + 7h - 7x] / h

y'(x) = lim(h - - > 0) : 7h/ h

y'(x) = 7

2.)

y = 5x³

y'(x) = lim(h - - > 0) : [y(x + h) - y(x)] / h

Obtain : y(x + h)

y(x) = 5x³

y(x + h) = 5(x + h)³

Substitute into formula :

y'(x) = lim(h - - > 0) : [5(x + h)³ - 5x³] / h

Expand :

y'(x) = lim(h - - > 0) : [5(x³ + 3x²h + 3xh² + h³) - 5x³] / h

y'(x) = lim(h - - > 0) : [5x³ + 15x²h + 15xh² + 5h³ - 5x³] / h

y'(x) = lim(h - - > 0) : [15x²h + 15xh² + 5h³] / h

y'(x) = lim(h - - > 0) : h(15x² + 15xh + 5h²) / h

y'(x) = lim(h - - > 0) : (15x² + 15xh + 5h²)

y'(x) = 15x²

answer: hi

explanation: