Balloon Calculus!

Given these pairs (either differentiating down or, adding a constant of integration, integrating up)...

... along with the chain rule...

... and the product rule...

... we illustrate some common results below...

Usually, you'll integrate sec² x by remembering that it's the derivative of tan x. If you want to forego that short-cut and integrate strategically, you can use the Weierstrass sub - see for example the integrate cosec section below.

Alternatively...

Alternatively, using the Weierstrass substitution, which is more round about but also a more widely applicable method...

Alternatively, using the Weierstrass substitution, which is more round about but also a more widely applicable method...

Examples of differentiating a product (3x² - 3) (x² - 2) / (x² + 2) x e^-kx relativisitic momentum x arctany d²/dt² e^t cos t √m ln m + e^m tan m (1 - e^-x2) (1 - e^-y2) 2x/(y² + 2) ∂/∂x, ∂/∂y... e^{x siny} f(x - y) x² e^-kx (5x³ + 1)⁸ (4x⁵ + 3)⁷ -8x^-x2 tanx ln|2x + 5| √t (1 - t²) x² e^-3e x² (x - 2)² 3x e^2x cos⁴(3t) sin(3t) 2x  ln |x² + 5| x(1 - x)^(3/5)

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... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (which is the inner function of the composite expression which is subject to the chain rule).

As with the product rule, differentiation is generally a clockwise journey...

Examples of differentiating a composite function ln(tanx) differentiate integral of composite directional derivatives sin³(cos²(3e^x2 )) ∂/∂x √ (x² + y²) f(4x³) 1 / (1 + 12.6e^-0.73x) Lorentz factor 3(2 - 8x)⁴ wave equation 3e^x2 sin(e^3x) sin²(sin(sinx)) tan²(x³) 5x³ - 3/(2 - 7x) + 3x⁵ - 15 sec³(π /2 - x) e^kx δ / δx e^xy √(x + √(x + √(x))) arcsin³(2x) [ sin( 1 + {cos( 1 + [tan(1+x)]⁴) }³) ]² [(4x + 3) / (5x - 2)]⁷ e^{sec x} (6x⁴ - 6x ^-3 - 2x + 5) ^-2 - ln | cos x | ln | 8x³ - 7x + 2.78 |

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Double composite partial derivative proof d/dx 4/3 x^{3/4 - x} Φ(u,v) = f(u^v cos v, u^v sin v z(u,v) = f(u² - v², 2uv) x = u(s,t), y = v(s,t) d/dx (x²)^ex d/dt (5 - 3t)^4t d/dx (tan x)^x d/dx x ^(1/x) d/dx f(x) ^g(x) d/dx x ^sinx d/dx 10x ^{ln x} ∂/∂x (x² - y²) / (4xy³)

Logarithmic differentiation d/dx cosx lnx / x² d/dx 4/3 x^{3/4 - x} y = 5x^{2x tan(6x)} d/dx x^7x

Implicit differentiation x² + wx + w² = 1 x²y³ + 2y = 6 x²y + 3x - 2y = 6 y + e^y = x - e^2x2 √y + x ln y = cos(xy²) 2x/(y² + 2) y + 2x siny = e^x d/dt √(x² + y²) x²y + xy² = 6 4x³ + 2xy - y³ = 1/2 x e^y = y - 1 e^x2 + y⁶ = 2x cos y x² + y² = 25 x³y / (1 - xy) 2x²y = sin(2x) x²y = x + y² sin3x cos2y = 6 3x² + 4xy³ = 9 6x + 10y dy/dx - 6 + 20 dy/dx x² - xy + ¾ x² = 7

Internal substitution differentiate an inverse

Differentiate a quotient

Note: the function 1/g would be written g/g² in the bottom row, just so we have a common denominator ready for simplifying the row... very like the standard quotient-rule formula. But, note also that if g is a power of a function (e.g. if it's sin²x) then we'll put sinx in the dashed balloon and write the second balloon in the bottom row as sinx over sin³x. This is (in these cases) a bit neater than the formula, which produces sin⁴x in the denominator.

Examples of differentiating a quotient (2x⁴ - 3x) / (4x - 1) 1/6 e^{2 sin x - 3x} / (3 + 2 cos x) (1 - x²)³ / (6 + 2x)³ (2x - 3)² / (x³ - 7)³ x (x²)/(x + 2) (1 - x²)/(x² + 1)² (x + 5)/(2x - 4) -4x/(x² - 1)² (1 + x²)/(1 + 2x²) 5 cosx/(4 - secx) x/(x + 2)² x²/e^-2x tan t (exponential form) ln|x| / √x √(7x - 2) / √(5x + 3) (4x + 3)⁷ / (5x - 2)⁷ 3x/(x² + 4) (4 - x²) / [ (x² + 4) ]² (1-s² ) / (1 + s² )²

Differentiate an inverse trig function

Examples of differentiating an inverse trig function d/dx arctan(y/x) d/dx arctan(4 - 5x) d/dt arcsec(2√t)

Similarly, to justify the general formula for differentiating an inverse: d/dx f  ^-1(x) = 1 / [f '(f  ^-1(x))]...

Integration by parts

Examples of integration by parts ∫ 5x sin(2x) dx ∫ secx(2x + x² tanx) dx ∫ x² e^4x dx ∫ e^x sinhx dx ∫ x / √(2x + 1) dx ∫ x⁹ √(x⁵ - 3) dx ∫ x³ √(2 + x²) dx ∫ 2x³ e^-x2 dx ∫ (1 - 2 cosx) (3 + 2 cosx) e^{(1/2 x + cosx} dx ∫ x² ln|4x| dx ∫ arctanx dx ∫ (5x⁴ + 4x⁵) / (x⁵ + x + 1) dx ∫ (x + 2) sin(x) dx          ∫ cos(2x)(5x + 1) dx ∫ u dv = uv - ∫ v du ∫ √x ln(x) dx ∫ x³ e^x2 dx ∫ 1 / [(x - 1)^3/4 (x + 2)^5/4] dx ∫ arcsinx / √(1 + x) dx ∫ 1 / [x² (x⁴ + 1)^3/4] dx ∫ 2x √(2x - 3) dx ∫ ln(2x)/x² dx ∫ 1/x e^1/x dx ∫ cos²x dx ∫ x / (2x + 1)⁴ dx ∫ sin^2 x dx ∫ ln(lnx) dx ∫ (lnx)²/x² dx ∫ √ (1 - x²) dx ∫ x arctanx dx ∫ x³ √ (x² + 1) dx ∫ sin²x / cos⁵x dx ∫ (1/3)u sinu du ∫ (1/5)u cos(u) du ∫ (2 sin x - 1) (sin x + 2) e^{2/3 sin x} dx ∫ √(a² - x²) dx ∫ (x + 2) (1 - e^-x/3) dx ∫ 5x e^4x dx ∫ ln (x² - x + 2) dx ∫ ln (x + 1) dx ∫ ln (2x + 1) dx

Parts twice ∫ √(x - √(x² - 4)) dx ∫ e^(2x) sinx dx ∫ e^x cosx dx ∫ 3x² e^x/2 dx ∫ cos(lnx) dx ∫ x² cos (π x) dx ∫ e^x sin(2x) dx ∫ x² sin(3x + 1) dx

Parts thrice ∫ (2x³ + 5x + 1) e^2x dx ∫ x³ sin x dx

Parts using triple product rule ∫ ₀¹ lnx ln(1 - x) dx

Recurrence formulae y_n = ∫ ₀¹ tⁿe^-t dt, n = 1, 2, 3,...

Integration by substitution / change of variable

Examples of integration by substitution ∫ (e^x - 1) / e^2x dx ∫ 1 / (64x² + 81) dx ∫ x / (1 + x²) dx ∫ (v - 1) / (v² + 2v + 2) dv ∫ x³ √(2 + x²) dx ∫ sin²(4t) cos(t) dt ∫ u dx ∫ 1/3x³ √(1 + x⁴) dx ∫ x / √(4 - x²) dx ∫ sinx / cos²x dx ∫ e^{- x} dx ∫ 1 / (tan 2x [1 - 2 ln(sin 2x)]) dx ∫ (e^x + 1) / (e^x - 1) dx ∫ x² √ (7 - 3x³) dx ∫ sec²(3x) dx ∫ √ x / ( 1 + x^1/4) dx ∫ arcsin √x / √ (x - x²) dx ∫ (5x² + 4)³ dx ∫ (lnx)²/x² dx ∫ 1 / (e^2x + 1) dx ∫ z³ e^-z2/2 dz ∫ x e^x2 dx ∫₀^a2/25 x²⁴ √(a² - x²⁵) dx ∫₀¹ (e^z + 21) / (e^z + 21z) dz ∫ 1 / (9 sin²x + 4 cos²x) dx ∫ 1 / [x (1 - ⁴√x)] dx ∫ 2x/(3 + x⁴) dx ∫ e^{sin x} sin(2x) dx ∫ x² √(2 + x³) dx ∫ p(p + 1)⁵ dp ∫ t/(t² + 2) dt ∫ t³ √(t² + 1) dt ∫ √x / (1 + x) dx ∫ 1 / (√x (1 + x)) dx ∫ 8(y - 9)^-3 dy ∫ 3x e^{2x2 - 1} dx ∫ e^{- x} dx ∫ sec²x / (4 + tan²x) dx ∫ sec x tan x dx ∫ 1 / [(a - x)(b - x)] dx ∫ ln | 1 - x | dx ∫ cos(a - x) dx ∫ tan⁵x sec⁴x dx ∫ cos⁴(3t) sin(3t) dt ∫₀^{π / 3} sin θ / cos² θ dθ ∫ 4t / (1 + 16t²) dt ∫ - 4x / (2x² + 1)² dx ∫ 6 sec²(5x) / [7 - 3 tan(5x)]⁷ dx ∫ ½ e^√x / √x dx ∫₁³ (x + 3) / (x² + 6x) dx ∫ ₀^π sin t / (1 + cos² t) dt ∫ y √(1 - y) dy

Integration by internal substitution

Examples of integration by internal substitution ∫ 1 / (1 + x²) dx ∫ 1 / [(x² + a²) √(x² + a² + b²) dx ∫ 1/(5 - 2√x) dx ∫ √(a² + x²) dx ∫₁² 1/√(4 - x²) dx ∫ (4 - x) / √(4 - x²) dx ∫ 6 / (9x² + 4) dx ∫ x³ √ (x - 3) dx ∫ √ x / ( 1 + x^1/4) dx ∫ 1/(x² + x + 1) dx ∫ 1/(3 + x²) dx ∫ √ (6 - 6x²) dx ∫ ₀¹ 1 / (1 + ³√ x) dx ∫ 1 / √ (x² + 25) dx ∫ 1/2 x² √ (1 + x²) dx ∫ √ (1 + 1/x²) dx          ∫ √ (1 + 4x²) dx          ∫ √ (1 - (x/a)²) dx ∫ 1 / (c² + x²)^(3/2) dy ∫ √(4 - x²) dx ∫ 1 / √(49 + x²) dx ∫ 1 / (x² + 1)² dx ∫ 1 / (x² √(x² - a²) dx ∫ 1 / √(1 - x²) dx

Weierstrass substitution ∫ 2 / (2 cosx + sinx) dx ∫ secx dx ∫ cosecx dx

Separation of variables

Get your equation in the form f(y) * dy/dx = f(x). (Not a complete separation). Fit the equation along the bottom row...

You can then extend the diagram on either side if necessary (chain-rule shapes shown for example).

Examples of separation of variables rate of evaporation from a dish of water x dy/dx - 3y + 2 = 0 dy/dx = -y / (4 + x) dP/dt = kP dF/dt = -2F 1/P dP/dt = k dy/dx = 2y (100 - t^2) dN/dt + 4t N = 0 dh/dt = (6 - h) /20 dy/dx = Cy

Integrating factor

If you can get your equation in the form      dy/dx + y f(x) = 0      or      dy/dx + y f(x) = g(x),

then fill just these places in the bottom row...

... and the others will follow...

Note: in place of y you can accomodate a function f(y)... as long as, in the place of dy/dx, you have d (f(y)) / dx.

Examples of choosing an integrating factor (x³ + y) dx - x dy = 0 birth and death rates e^x dy/dx = 1 - 2e^xy dy/dx = 2x (1 + x² - y) dy/dx = y/x + y²/x² x dy/dx - 4y = 0 dθ/dt + θ/10 = 5 - 5t/2 x² y' + xy = x e^x dy/dx - y / (x + 1) = x - 1

Exact equations

If you can get the left-side of your equation to fit the lower level of the product rule without need of an integrating factor (see above), the equation is exact. Most examples of 'exact equations', however, don't necessarily fit the product rule as such, but do fit the multi-variable chain rule...

... of which the product rule is a special case. To be able to find F you need an equation that fits the bottom row and also passes the 'exactness test'. That is, you need to check that the partial derivative of M with respect to y equals that of N with respect to x. That settled, you know you can integrate as follows...

More examples of solving an exact equation 3x²/y² + y²/x² - 2/x³ 6/y⁴ - 2/x³ + (6/y⁴ - 2x³/y³ - 2y/x) y' = 0 ∂u/∂x . ∂u/∂y = xy F(x,y,z) = (yz² - 2)i + (xz² - e^2x - 6y²)j + (2xyz + 4z - 2y e^2x)k