Differentiate the following functions using first principles

Question

a)f(x)=4x²+x
b)g(x)=-8x²-5x+13
c)m(y)=1/1-yx
d) f(x)=-8/y
e) f(x)=square root x

Answer 1

a) f'(x) = lim(h->0) [f(x+h)-f(x)]/h
= lim(h->0) [4(x+h)²+(x+h)-(4x²+x)]/h
= lim(h->0) [4x²+8xh+4h²+x+h-4x²-x]/h
= lim(h->0) [8xh+4h²+h]/h
= lim(h->0) 8x+4h+1
= 8x+1

So, f'(x) = 8x+1

b) g'(x) = lim(h->0) [g(x+h)-g(x)]/h
= lim(h->0) [-8(x+h)²-5(x+h)+13+8x²+5x-13]/h
= lim(h->0) [-8x²-16xh-8h²-5x-5h+8x²+5x-13]/h
= lim(h->0) [-16xh-8h²-5h]/h
= lim(h->0) -16x-8h-5
= -16x-5

So, g'(x) = -16x-5

c) m'(y) = lim(h->0) [m(y+h)-m(y)]/h
= lim(h->0) [(1/(1-(y+h)x))-(1/(1-yx))]/h
= lim(h->0) [(1-yx)-(1-(y+h)x)]/[h(1-(y+h)x)(1-yx)]
= lim(h->0) [yx-yh-yx+hx]/[h(1-(y+h)x)(1-yx)]
= lim(h->0) [-yh+hx]/[h(1-(y+h)x)(1-yx)]
= lim(h->0) [-y+x]/[(1-(y+h)x)(1-yx)]
= -(y/x²)

So, m'(y) = -(y/x²)

d) f'(x) = lim(h->0) [f(x+h)-f(x)]/h
= lim(h->0) [-8/(y+h)+8/y]/h
= lim(h->0) [-8(y-(y+h))/hy(h+y)]
= lim(h->0) 8/(y²+yh)
= 8/y²

So, f'(x) = 8/y²

e) f'(x) = lim(h->0) [f(x+h)-f(x)]/h
= lim(h->0) [(√(x+h)-√(x))/h]
= lim(h->0) [(√(x+h)-√(x))/h] * [(√(x+h)+√(x)+h)/(√(x+h)+√(x)+h)]
= lim(h->0) [(x+h-x)/(h(√(x+h)+√(x)+h))]
= lim(h->0) 1/(√(x+h)+√(x)+h)
= 1/(2√x)

So, f'(x) = 1/(2√x)

Answer 2

Integrate sin² x do

Answer 3

We can use the trigonometric identity: sin²x = (1-cos(2x))/2

So, ∫sin²x dx = ∫(1-cos(2x))/2 dx

= 1/2 ∫1 dx - 1/2 ∫cos(2x) dx

= 1/2 x - 1/4 sin(2x) + C

where C is the constant of integration.