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Math  /  Calculus

Question1. Find the derivatives of the following functions: a) y=x3sinh⁡xy=\frac{x^{3}}{\sinh x} b) y=tanh⁡32xy=\tanh ^{3} 2 x c) y=ecosh⁡4x2y=e^{\cosh 4 x^{2}}

Studdy Solution

STEP 1

1. We are given three functions for which we need to find the derivatives.
2. We will use standard differentiation rules, including the quotient rule, chain rule, and product rule as necessary.

STEP 2

1. Differentiate y=x3sinh⁡x y=\frac{x^{3}}{\sinh x} using the quotient rule.
2. Differentiate y=tanh⁡32x y=\tanh ^{3} 2 x using the chain rule.
3. Differentiate y=ecosh⁡4x2 y=e^{\cosh 4 x^{2}} using the chain rule.

STEP 3

Differentiate y=x3sinh⁡x y=\frac{x^{3}}{\sinh x} using the quotient rule. The quotient rule states:
(uv)′=u′v−uv′v2 \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}
where u=x3 u = x^3 and v=sinh⁡x v = \sinh x .

STEP 4

Find the derivatives of u u and v v :
u′=ddx(x3)=3x2 u' = \frac{d}{dx}(x^3) = 3x^2 v′=ddx(sinh⁡x)=cosh⁡x v' = \frac{d}{dx}(\sinh x) = \cosh x

STEP 5

Apply the quotient rule:
y′=3x2sinh⁡x−x3cosh⁡x(sinh⁡x)2 y' = \frac{3x^2 \sinh x - x^3 \cosh x}{(\sinh x)^2}

STEP 6

Differentiate y=tanh⁡32x y=\tanh ^{3} 2 x using the chain rule. Let u=tanh⁡2x u = \tanh 2x and y=u3 y = u^3 .

STEP 7

Find the derivative of y=u3 y = u^3 with respect to u u :
dydu=3u2 \frac{dy}{du} = 3u^2

STEP 8

Find the derivative of u=tanh⁡2x u = \tanh 2x with respect to x x :
\frac{du}{dx} = \frac{d}{dx}(\tanh 2x) = 2 \sech^2 2x

STEP 9

Apply the chain rule:
y' = \frac{dy}{du} \cdot \frac{du}{dx} = 3(\tanh 2x)^2 \cdot 2 \sech^2 2x y' = 6 (\tanh 2x)^2 \sech^2 2x

STEP 10

Differentiate y=ecosh⁡4x2 y=e^{\cosh 4 x^{2}} using the chain rule. Let u=cosh⁡4x2 u = \cosh 4x^2 and y=eu y = e^u .

STEP 11

Find the derivative of y=eu y = e^u with respect to u u :
dydu=eu \frac{dy}{du} = e^u

STEP 12

Find the derivative of u=cosh⁡4x2 u = \cosh 4x^2 with respect to x x :
dudx=ddx(cosh⁡4x2)=sinh⁡4x2⋅ddx(4x2) \frac{du}{dx} = \frac{d}{dx}(\cosh 4x^2) = \sinh 4x^2 \cdot \frac{d}{dx}(4x^2) dudx=sinh⁡4x2⋅8x \frac{du}{dx} = \sinh 4x^2 \cdot 8x

STEP 13

Apply the chain rule:
y′=dydu⋅dudx=ecosh⁡4x2⋅sinh⁡4x2⋅8x y' = \frac{dy}{du} \cdot \frac{du}{dx} = e^{\cosh 4x^2} \cdot \sinh 4x^2 \cdot 8x y′=8xecosh⁡4x2sinh⁡4x2 y' = 8x e^{\cosh 4x^2} \sinh 4x^2
The derivatives of the functions are: a) y′=3x2sinh⁡x−x3cosh⁡x(sinh⁡x)2 y' = \frac{3x^2 \sinh x - x^3 \cosh x}{(\sinh x)^2} b) y' = 6 (\tanh 2x)^2 \sech^2 2x c) y′=8xecosh⁡4x2sinh⁡4x2 y' = 8x e^{\cosh 4x^2} \sinh 4x^2

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