Answer:
y'(x) = 7 ;
y'(x) = 15x²
Step-by-step 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