What is the second derivative for #y(x)=5e^(pi x)+cos (y(x))# ?

Question

Paisley Johnson · Accepted Answer

#y''=(5pi^2e^(pix) ((1+siny)^2-5e^(pix)cosy))/(1+siny)^3# #d/dx(y(x))=d/dx(5e^(pix)+cos(y(x))# #y'=5pie^(pix)-siny*y'# #y'=(5pie^(pix))/(1+siny)# #d/dx(y')=(5pi^2e^(pix)*(1+siny)-5pie^(pix)*cosy*y')/(1+siny)^2# #y''=(5pi^2e^(pix)*(1+siny)-5pie^(pix)*cosy (5pie^(pix))/(1+siny))/(1+siny)^2# #y''=(5pi^2e^(pix)*(1+siny)-5pie^(pix)*cosy (5pie^(pix))/(1+siny))/(1+siny)^2# #y''=(5pi^2e^(pix)*(1+siny)^2-25pi^2e^(2pix)*cosy)/(1+siny)^3# #y''=(5pi^2e^(pix) ((1+siny)^2-5e^(pix)cosy))/(1+siny)^3#

Evelyn Agard · Answer

undefined To find the second derivative of the function $ y(x) = 5e^{\pi x} + \cos(y(x)) $, you first need to find the first derivative with respect to $ x $, then take the derivative of that result again with respect to $ x $.

First derivative:
$$ y'(x) = 5\pi e^{\pi x} - \sin(y(x)) \cdot y'(x) $$

Solve for $ y'(x) $:
$$ y'(x) + \sin(y(x)) \cdot y'(x) = 5\pi e^{\pi x} $$
$$ y'(x) (1 + \sin(y(x))) = 5\pi e^{\pi x} $$
$$ y'(x) = \frac{5\pi e^{\pi x}}{1 + \sin(y(x))} $$

Now, take the derivative of $ y'(x) $ with respect to $ x $ to find the second derivative:

$$ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{5\pi e^{\pi x}}{1 + \sin(y(x))} \right) $$

Using the quotient rule:

$$ \frac{d^2y}{dx^2} = \frac{(5\pi e^{\pi x})'(1 + \sin(y(x))) - 5\pi e^{\pi x}(1 + \sin(y(x)))'}{(1 + \sin(y(x)))^2} $$

$$ = \frac{(5\pi^2 e^{\pi x})(1 + \sin(y(x))) - 5\pi e^{\pi x} \cos(y(x)) \cdot y'(x)}{(1 + \sin(y(x)))^2} $$

Substitute $ y'(x) $ into the equation:

$$ = \frac{(5\pi^2 e^{\pi x})(1 + \sin(y(x))) - 5\pi e^{\pi x} \cos(y(x)) \cdot \frac{5\pi e^{\pi x}}{1 + \sin(y(x))}}{(1 + \sin(y(x)))^2} $$

$$ = \frac{(5\pi^2 e^{\pi x})(1 + \sin(y(x))) - (25\pi^2 e^{2\pi x}) \cos(y(x))}{(1 + \sin(y(x)))^2} $$