A ray that contains a point that is equidistant from both sides of the angle. *

Mathematics

1 answer:

seraphim [82]3 years ago

3 0

Consider a angle â BAC and the point D on its defector Assume that DB is perpendicular to AB and DC is perpendicular to AC. Lets prove DB and DC are congruent (that is point D is equidistant from sides of an angle â BAC Proof Consider triangles Î”ADB and Î”ADC Both are right angle, â ABD= â ACD=90 degree They have congruent acute angle â BAD and â CAD( since AD is angle bisector) They share hypotenuse AD therefore these right angle are congruent by two angle and sides and, therefore, their sides DB and DC are congruent too, as luing across congruent angles

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Differentiating a Logarithmic Function In Exercise, find the derivative of the function.

sp2606 [1]

Answer:

$\dfrac{d(f(x))}{dx} = \dfrac{2}{x}$

Step-by-step explanation:

We are given the following function in the question:

$f(x) = \ln (3x^2)$

We have to derivate the given function.

Formula:

$\dfrac{d(\ln x)}{dx} = \dfrac{1}{x}\\\\\dfrac{d(x^n)}{dx} = nx^{n-1}$

The derivation takes place in the following manner

$f(x) = \ln (3x^2)\\\\\dfrac{d(f(x))}{dx} = \displaystyle\frac{d(\ln(3x^2))}{dx}\\\\=\frac{1}{3x^2}\times \frac{d(3x^2)}{dx}\\\\= \frac{1}{3x^2}\times (6x)\\\\=\frac{6x}{3x^2}\\\\=\frac{2}{x}$