Question

Find the partial derivative of the function f(x,y)=Integral of cos(-7t^2-6t-1)dt. Find fx(x,y) and fy(x,y)

Answer 1

Answer:

$\mathbf{\dfrac{\partial f}{\partial x}= cos (-7x^2 -6x - 1)}$

$\mathbf{\dfrac{\partial f}{\partial y}= cos ( -7y^2 -6y-1)}$

Step-by-step explanation:

Given that :

$f(x,y) = \int ^x_y cos (-7t^2 -6t-1) dt$

Using the Leibnitz rule of differentiation,

$\dfrac{d}{dt} \int ^{b(t)}_{a(t)} f(x,t) dt= f(b(t),t) *b'(t) -f(a(t),t) * a' (t) + \int^{b(t)}_{a(t)} \dfrac{\partial f}{\partial t} \ dt$

To find: fx(x,y)

$\dfrac{\partial f}{\partial x}= \dfrac{\partial }{\partial x} [ \int ^x_y cos (-7t^2 -6t -1 ) \ dt]$

$\dfrac{\partial f}{\partial x}= \dfrac{\partial x}{\partial x} cos (-7x^2 -6x -1 ) - \dfrac{\partial y}{\partial x} * cos (-7y^2 -6y-1) + \int ^x_y [\dfrac{\partial }{\partial x} \ \{cos (-7t^2-6t-1)\}] \ dt$

$\dfrac{\partial f}{\partial x}= cos (-7x^2 -6x - 1) -0+0$

$\mathbf{\dfrac{\partial f}{\partial x}= cos (-7x^2 -6x - 1)}$

To find: fy(x,y)

$\dfrac{\partial f}{\partial y}= \dfrac{\partial }{\partial y} [ \int ^x_y cos (-7t^2 -6t -1 ) \ dt]$

$\dfrac{\partial f}{\partial y}= \dfrac{\partial x}{\partial y} cos (-7x^2 -6x -1 ) - \dfrac{\partial y}{\partial y} * cos (-7y^2 -6y-1) + \int ^x_y [\dfrac{\partial }{\partial y} \ \{cos (-7t^2-6t-1)\}] \ dt$

$\dfrac{\partial f}{\partial y}= 0 - cos ( -7y^2 -6y-1)+0$

$\mathbf{\dfrac{\partial f}{\partial y}= cos ( -7y^2 -6y-1)}$