May 15, 2024
lim_(x→3) [(x^2 - 9)/(x - 3)]:(x^2 - 9) = (x + 3)(x - 3).(x - 3), resulting in lim_(x→3) [(x + 3)].3 + 5 / 3 + 3 = 8 / 6 = 4 / 3.d/dx [x^n] = nx^(n-1).x^4: Result is 4x^3.x^(-1): Result is -x^(-2) reversed for reciprocal.6x^5 - 3/x^2 + 1/(2√x).f(x) to be continuous at x=c, both pieces of the function must be equal at x=c.2cx - 6 = x^2 + cx at x=3.2c(3) - 6 = 9 + 3c -> 6c - 6 = 9 + 3c -> 3c = 15 -> c = 5.e^(u) derivative is e^(u) * u'.ln(u) derivative is u'/u.(f*g)' = f'*g + f*g'.e^(4x) * ln(2x+5):4e^(4x)(ln(2x+5)) + e^(4x)*(2/(2x + 5)).(x^n dx) = (x^(n+1))/(n+1) + C.(4x^5 + x^4 - 3x^2)/x^2(4x^3 + x^2 - 3x).x^4/4 + (x^3/3) - (3x^2/2) + C.y - y1 = m(x - x1), where m is slope.2x^3 + 4xy^2 + y^3 = 107 at (2,3).16x + 25y = 107.f'(x) = lim_(h→0) [f(x+h) - f(x)] / h.cos x._2x√(3x^2 + 5) dx.u = 3x^2 + 5, then, du = 6x dx or dx = du/(6x).2/3(u^(3/2)) + C.V=πr^2h).dh/dt = 3 ft/min, r=3ft, find dV/dt.dV/dt = π * 9 * 3 = 27π ft³/min.-x²+16x+5 has max point found by solving f'(x)=0.1/(b-a) ∫_[a to b] f(x) dx.f(x) = x^3 + 8x - 4 from [1, 5].59 for average value._d/dx f(g(x)) = f'(g(x)) * g'(x).(2x^3 - 7x^2)^8.
8(2x^3 - 7x^2)^7 * (6x^2 - 14x) simplified.limit involving fractions simplifies to cancellation method.