Calculus Solutions, Important Examples, Formulas and Videos

Differentiability

Derivatives of composite function:

Example: find the derivative of f, where

Solution: \frac{d}{dx}f(x)= \frac{d}{dx}\left [ (2x+1)^{3}\right ]

= \frac{d}{dx}(8x^{3}+12x^{3}+6x+1)

= 24x^{2}+24x+6

= 6(2x+1)^{2}

Now, observe that f(x)= (h\:o\:g)(x)

Where g(x) =2x+1 , and h(x)=x^{3} .put t = g(x)=2x +1.Then f (x) = h(t) = t^{3} .Thus

\frac{\mathrm{d} f}{\mathrm{d} x}=6(2x+1)^{2}=3(2x+1)^{2}\cdot 2=3t^{2}\cdot 2=\frac{\mathrm{d} h}{\mathrm{d} t}\cdot \frac{\mathrm{d} t}{\mathrm{d} x}

Derivatives of Functions in Parametric Forms: Some times the relation between two variables is neither explicit nor implic, but some link of a third variable with each of the two variables, separately , establishes a relation between the first two variables. In such a situation, we say that the relation between them is expressed via a third variable. The third variable is called the parameter . More precisely, a relation expressed between two variables x and y in the form x = f(t), y= g(t) is said to be parametric form with t as a parameter.

In order to find derivative of function in s uch form, we have by chain rule:
\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t}\cdot \frac{\mathrm{d} t}{\mathrm{d} x}

Thus , \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{g^{'}(t)}{f^{'}(t)}=\left ( as\: \frac{\mathrm{d} y}{\mathrm{d} x} =g^{'}(t)\: and\: \frac{\mathrm{d} x}{\mathrm{d} t}=f^{'}(t)\right )

Points to be Remember :

  • A real valued function is continuous at a point in its domain if the limit of the function at that point equals the value of the function at that point. A function is continuous if it is continuous on the whole of its domain .
  • Sum, difference, product and quotient of continuous functions are continuous. i.e., if f and g are continuous functions, then
  • (f\pm g)(x)=f(x)\pm g(x) is continuous .
    (f. g) (x) = f(x) . g(x) is continuous
    \left (\frac{f}{g}\right )(x)=\frac{f(x)}{g(x)} (where g(x) ≠ 0) is continuous ) .
  • Every differentiable function is continuous, but the converse is not true.
  • Chain rule is rule to differentiate composites of functions. If f = v o u , t = u(x) and if both
    \frac{\mathrm{d} t}{\mathrm{d} x}\: and\: \frac{\mathrm{d} v}{\mathrm{d} t} exist then
    \frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\cdot \frac{\mathrm{d} t}{\mathrm{d} x}
  • Following are some of the standard derivatives (in appropriate domains):
    \frac{\mathrm{d} }{\mathrm{d} x}(\sin^{-1})=\frac{1}{\sqrt{1-x^{2}}}
    \frac{\mathrm{d} }{\mathrm{d} x}(\cos^{-1})=\frac{1}{\sqrt{1-x^{2}}}
    \frac{\mathrm{d} }{\mathrm{d} x}(\tan^{-1})=\frac{1}{{1+x^{2}}}
    \frac{\mathrm{d} }{\mathrm{d} x}(\cot^{-1})=\frac{-1}{{1+x^{2}}}
    \frac{\mathrm{d} }{\mathrm{d} x}(\sec^{-1})=\frac{1}{x\sqrt{1-x^{2}}}
    \frac{\mathrm{d} }{\mathrm{d} x}(\cos ec^{-1})=\frac{-1}{x\sqrt{1-x^{2}}}
    \frac{\mathrm{d} }{\mathrm{d} x}(e^{x})=e^{x}
    \frac{\mathrm{d} }{\mathrm{d} x}(logx)=\frac{1}{x}
  • Logarithmic differentiation is a powerful technique to differentiate functions of the form  f(x)=\left [ u(x) \right ]^{v(x)} Here both f (x) and u(x) need to be positive for this technique to make sense.
  • Rolle’s Theorem : If f: [a, b] →R is continuous on [a, b] and differentiable on (a, b) such that f(a) = f(b), then There exists some c in (a, b) such that f′(c) = 0.
  • Mean Value Theorem: If f: [a, b] →R is continuous on [a, b] and differentiable on (a, b). Then there exists some c in (a, b) such that
    f^{'}(c)=\frac{f(b)-f(a)}{b-a}
  • Second derivatives test : Let f be a function defined on an intervalI and c ∈ I. Let f be twice differentiable at c. Then
    (i) x = c is a point of local maxima if f′(c) = 0 and f′′(c) < 0The values f(c) is local maximum value of f . (ii) x = c is a point of local minima if f′(c) = 0 and f′′(c) > 0 In this case, f (c) is local minimum value of f .
    (iii) The test fails if f′(c) = 0 and f′′(c) = 0.In this case, we go back to the first derivative test and find whether
    C is a point of maxima, minima or a point of inflexion.
  • Working rule for finding absolute maxims and/or absolute minims
    Step 1: Find all critical points of f in the interval, i.e., find points x where either f′(x) = 0 or f is not differentiable.
    Step 2 : Take the end points of the interval.
    Step 3 : At all these points (listed in Step 1 and 2), calculate the values of f.
    Step 4 : Identify the maximum and minimum values of f out of the values calculated in Step 3. This maximum value will be the absolute maximum value of f and the minimum value will be the absolute minimum value of f .

 

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