Mathematics, 29.01.2020 10:46 springtrap77391

What is the first derivative of 3secx?

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——————————

Given the function

y = 3 · sec(x)

   dy
find —— .
   dx

Using the definition of derivative:

$\mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}~\dfrac{y(x+h)-y(x)}{h}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}~\dfrac{3\,sec(x+h)-3\,sec(x)}{h}}\\\\\\\ \mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}~3\cdot \big[sec(x+h)-sec(x)\big]\cdot \dfrac{1}{h}}$

Secant is the reciprocal of cosine:

$\mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}~3\cdot \left[\dfrac{1}{cos(x+h)}-\dfrac{1}{cos(x)}\right]\cdot \dfrac{1}{h}}$

Reduce those fractions in brackets to the same denominator:

$\mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}~3\cdot \left[\dfrac{cos(x)}{cos(x+h)\cdot cos(x)}-\dfrac{cos(x+h)}{cos(x+h)\cdot cos(x)}\right]\cdot \dfrac{1}{h}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}~3\cdot \dfrac{cos(x)-cos(x+h)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{1}{h}}$

Recall one trogonometric identity (cosine of a sum):

• cos(x + h) = cos(x) · cos(h) – sin(x) · sin(h)

then you have

$\mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}~3\cdot \dfrac{cos(x)-\big[cos(x)\cdot cos(h)-sin(x)\cdot sin(h)\big]}{cos(x+h)\cdot cos(x)}\cdot \dfrac{1}{h}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}~3\cdot \dfrac{cos(x)-cos(x)\cdot cos(h)+sin(x)\cdot sin(h)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{1}{h}}$

Take out the common factor cos(x) from the first two terms in that numerator:

$\mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}~3\cdot \dfrac{cos(x)\cdot \big[1-cos(h)\big]+sin(x)\cdot sin(h)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{1}{h}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}~3\cdot \dfrac{cos(x)\cdot \big[1-cos(h)\big]+sin(x)\cdot sin(h)}{cos(x+h)\cdot cos(x)\cdot h}}$

Now, split up that first fraction into a sum of two fractions:

$\mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}\left[3\cdot \dfrac{cos(x)\cdot \big[1-cos(h)\big]}{cos(x+h)\cdot cos(x)\cdot h}+3\cdot \dfrac{sin(x)\cdot sin(h)}{cos(x+h)\cdot cos(x)\cdot h}\right]}\\\\\\ \mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}\left[\dfrac{3\,cos(x)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{1-cos(h)}{h}+\dfrac{3\,sin(x)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{sin(h)}{h}\right]}$

Multiply and divide that first term in that sum by $\mathsf{\big[1+cos(h)\big]:}$

$\mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}\left[\dfrac{3\,cos(x)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{\big[1-cos(h)\big]\cdot \big[1+cos(h)\big]}{h\cdot \big[1+cos(h)\big]}+\dfrac{3\,sin(x)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{sin(h)}{h}\right]}\\\\\\ \mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}\left[\dfrac{3\,cos(x)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{1-cos^2(h)}{h\cdot \big[1+cos(h)\big]}+\dfrac{3\,sin(x)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{sin(h)}{h}\right]}$

But 1 – cos²(h) = sin²(h):

$\mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}\left[\dfrac{3\,cos(x)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{sin^2(h)}{h\cdot \big[1+cos(h)\big]}+\dfrac{3\,sin(x)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{sin(h)}{h}\right]}\\\\\\ \mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}\left[\dfrac{3\,cos(x)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{sin(h)}{1+cos(h)}\cdot \dfrac{sin(h)}{h}+\dfrac{3\,sin(x)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{sin(h)}{h}\right]}$

sin(h)
Take out the common factor ———— :
         h

$\\\mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}\left[\dfrac{3\,cos(x)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{sin(h)}{1+cos(h)}+\dfrac{3\,sin(x)}{cos(x+h)\cdot cos(x)}\right]\cdot \dfrac{sin(h)}{h}}\\\\\\ \vdots\qquad\textsf{(applying limit properties)}\\\\\\ \mathsf{\dfrac{dy}{dx}=\underset{h\to 0}{\ell im}\left[\dfrac{3\,cos(x)}{cos(x+h)\cdot cos(x)}\cdot \dfrac{sin(h)}{1+cos(h)}+\dfrac{3\,sin(x)}{cos(x+h)\cdot cos(x)}\right]\cdot \underset{h\to 0}{\ell im}~\dfrac{sin(h)}{h}}$

The limit of sin(h)/h as h approaches 0 equals 1 (fundamental trigonometric limit). So the derivative is

$\mathsf{\dfrac{dy}{dx}=\left[\dfrac{3\,cos(x)}{cos(x+0)\cdot cos(x)}\cdot \dfrac{sin(0)}{1+cos(0)}+\dfrac{3\,sin(x)}{cos(x+0)\cdot cos(x)}\right]\cdot 1}\\\\\\ \mathsf{\dfrac{dy}{dx}=\left[\dfrac{3\,cos(x)}{cos(x)\cdot cos(x)}\cdot \dfrac{0}{1+1}+\dfrac{3\,sin(x)}{cos(x)\cdot cos(x)}\right]\cdot 1}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{3\,sin(x)}{cos^2(x)}}\\\\\\ \mathsf{\dfrac{dy}{dx}=3\cdot \dfrac{1}{cos(x)}\cdot \dfrac{sin(x)}{cos(x)}}$

∴ $\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=3\cdot sec(x)\cdot tan(x)}\end{array}}$ ✔

I hope this helps. =)

Tags: derivative function definition limit secant trigonometric trig differential integral calculus

Answer from: Quest

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Answer from: Quest

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