Discover the function's initial partial derivatives.&nbsp;f(x,y)=ax+bycx+dy

Question

Discover the function&#039;s initial partial derivatives.&amp;nbsp;f(x,y)=ax+bycx+dy

hosentak · Accepted Answer

Found the answer:

Jeffrey Jordon · Accepted Answer

Consider about the next function:f(x,y)=ax+bycx+dyDiscover the function&#039;s initial partial derivativesdfdx=ddx(ax+bycx+dy)=(cx+dy)ddx(ax+by)−(ax+by)ddx(cx+dy)(cx+dy)2=(cx+dy)(a)−(ax+by)(c)(cx+dy)2=acx+ady−acx−bcy(cx+dy)2=ady−bcy(cx+dy)2=(ad−bc)y(cx+dy)2Therefore,&amp;nbsp;dfdx=(ad−bc)y(cx+dy)2dfdx=ddy(ax+bycx+dy)=(cx+dy)(b)−(ax+by)(d)(cx+dy)2=bcx+bdy−adx−bdy(cx+dy)2=bcx−adx(cx+dy)2=(bc−ad)x(cx+dy)2Therefore,&amp;nbsp;dfdx=(bc−ad)x(cx+dy)2

xleb123 · Accepted Answer

Using the quotient rule, we have:∂f∂x=(c·(cx+dy)−(ax+by)·c)(cx+dy)2=c(cx+dy)−acx−bcy(cx+dy)2∂f∂y=(d·(cx+dy)−(ax+by)·d)(cx+dy)2=d(cx+dy)−adx−bdy(cx+dy)2Thus, the partial derivatives are:∂f∂x=c(cx+dy)−acx−bcy(cx+dy)2∂f∂y=d(cx+dy)−adx−bdy(cx+dy)2

Andre BalkonE · Accepted Answer

Step 1: First, let&#039;s find the partial derivative with respect to x (∂f∂x):∂f∂x=∂∂x(ax+bycx+dy)To differentiate this expression, we can use the quotient rule. The quotient rule states that for functions u(x) and v(x), the derivative of u(x)v(x) is given by:ddx(u(x)v(x))=u′(x)·v(x)−u(x)·v′(x)(v(x))2Using the quotient rule on our function, we have:∂f∂x=(a·(cx+dy)−(ax+by)·c)·(cx+dy)(cx+dy)2Simplifying this expression further, we get:∂f∂x=acx2+adxy+acy2−acx2−bcxy−bcy2(cx+dy)2Canceling out the common terms, we obtain:∂f∂x=adxy−bcxy(cx+dy)2Step 2: Next, let&#039;s find the partial derivative with respect to y (∂f∂y):∂f∂y=∂∂y(ax+bycx+dy)Using the quotient rule again, we have:∂f∂y=(a·(cx+dy)−(ax+by)·d)·(cx+dy)(cx+dy)2Simplifying this expression further, we get:∂f∂y=acxy+ady2+bcy2−adxy−bdy2(cx+dy)2Canceling out the common terms, we obtain:∂f∂y=acxy−bdy2(cx+dy)2Hence, the initial partial derivatives of the function f(x,y)=ax+bycx+dy are:∂f∂x=adxy−bcxy(cx+dy)2and∂f∂y=acxy−bdy2(cx+dy)2Note: In the above expressions, a, b, c, and d are constants.

fudzisako · Accepted Answer

To discover the initial partial derivatives of the function f(x,y)=ax+bycx+dy, we need to calculate the partial derivatives with respect to x and y. Let&#039;s start by finding the partial derivative with respect to x, denoted as ∂f∂x:∂f∂x=∂∂x(ax+bycx+dy)Using the quotient rule, we have:∂f∂x=(cx+dy)∂∂x(ax+by)−(ax+by)∂∂x(cx+dy)(cx+dy)2Now, let&#039;s calculate the individual derivatives. The derivative of ax+by with respect to x is a since b is a constant:∂∂x(ax+by)=aSimilarly, the derivative of cx+dy with respect to x is c since d is a constant:∂∂x(cx+dy)=cSubstituting these values back into the equation, we get:∂f∂x=(cx+dy)a−(ax+by)c(cx+dy)2Simplifying the numerator, we have:∂f∂x=acx+ady−acx−bcy(cx+dy)2This further simplifies to:∂f∂x=ady−bcy(cx+dy)2Now, let&#039;s calculate the partial derivative with respect to y, denoted as ∂f∂y:∂f∂y=∂∂y(ax+bycx+dy)Using the quotient rule again, we have:∂f∂y=(cx+dy)∂∂y(ax+by)−(ax+by)∂∂y(cx+dy)(cx+dy)2The derivative of ax+by with respect to y is b since a is a constant:∂∂y(ax+by)=bSimilarly, the derivative of cx+dy with respect to y is d since c is a constant:∂∂y(cx+dy)=dSubstituting these values back into the equation, we get:∂f∂y=(cx+dy)b−(ax+by)d(cx+dy)2Simplifying the numerator, we have:∂f∂y=bcx+bdy−adx−bdy(cx+dy)2This further simplifies to:∂f∂y=bcx−adx(cx+dy)2In summary, the initial partial derivatives of the function f(x,y)=ax+bycx+dy are:∂f∂x=ady−bcy(cx+dy)2∂f∂y=bcx−adx(cx+dy)2

find the first partial derivatives of the function f(x,y)=\frac{ax+by}{cx+dy}

Answered question

Answer & Explanation

New Questions in Differential Calculus