Find \(\displaystyle \frac{dz}{dt}\) if \(\displaystyle z=xln(y)\), \(\displaystyle x=sin(t)\) and \(\displaystyle y=cos(t)\).

Keep in mind, when taking the derivative with respect to \(\displaystyle x\), \(\displaystyle y\) is treated as a constant, and when taking the derivative with respect to \(\displaystyle y\), \(\displaystyle x\) is treated as a constant.

\(\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\)

\(\displaystyle \frac{dz}{dt}=(ln(y))(cos(t))+(x\frac{1}{y})(-sin(t))\) 

\(\displaystyle \frac{dz}{dt}=ln(y)cos(t)-\frac{x}{y}sin(t)\)

To put \(\displaystyle \frac{dz}{dt}\) solely in terms of \(\displaystyle x\) and  \(\displaystyle y\), we substitute the definitions of \(\displaystyle x\) and \(\displaystyle y\)  given in the question,  \(\displaystyle x=sin(t)\) and \(\displaystyle y=cos(t)\).

\(\displaystyle \frac{dz}{dt}=ln(cos(t))cos(t)-\frac{sin(t)}{cos(t)}sin(t)\)

\(\displaystyle \frac{dz}{dt}=ln[cos(t)]cos(t)-\frac{sin^2(t)}{cos(t)}\)

Find \(\displaystyle \frac{dz}{dt}\) if \(\displaystyle z=x^2y-y^2x\), \(\displaystyle x=t^2-5t\) and \(\displaystyle y=3t+4\).

Keep in mind, when taking the derivative with respect to \(\displaystyle x\), \(\displaystyle y\) is treated as a constant, and when taking the derivative with respect to \(\displaystyle y\), \(\displaystyle x\) is treated as a constant.

\(\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\)

\(\displaystyle \frac{dz}{dt}=(2xy-y^2)(2t-5)+(x^2-2yx)(3)\)

\(\displaystyle \frac{dz}{dt}=(2xy-y^2)(2t-5)+3(x^2-2xy)\)

To put \(\displaystyle \frac{dz}{dt}\) solely in terms of \(\displaystyle x\) and  \(\displaystyle y\), we substitute the definitions of \(\displaystyle x\) and \(\displaystyle y\)  given in the question,  \(\displaystyle x=t^2-5t\) and \(\displaystyle y=3t+4\).

\(\displaystyle \frac{dz}{dt}=(2(t^2-5t)(3t+4)-(3t+4)^2)(2t-5)+3((t^2-5t)^2-2(t^2-5t)(3t+4))\)

\(\displaystyle \frac{dz}{dt}=[2(3t^3-15t^2+4t^2-20t)-(9t^2+24t+16)](2t-5) +3[(t^4-10t^3+25t^2)-2(3t^3-15t^2+4t^2-20t)]\)

\(\displaystyle \frac{dz}{dt}=[6t^3-30t^2+8t^2-40t-9t^2-24t-16](2t-5) +3[t^4-10t^3+25t^2-6t^3+30t^2-8t^2+40t]\)

\(\displaystyle \frac{dz}{dt}=[6t^3-31t^2-64t-16](2t-5) +3[t^4-16t^3+47t^2+40t]\)

\(\displaystyle \frac{dz}{dt}=12t^4-62t^3-128t^2-32t -30t^3+155t^2+320t+80 +3t^4-48t^3+141t^2+120t\)

\(\displaystyle \frac{dz}{dt}=15t^4 -140t^3 +168t^2 +408t +80\)

Find \(\displaystyle \frac{dz}{dt}\) if \(\displaystyle z=e^{5xy}\), \(\displaystyle x=\sqrt{t}\) and \(\displaystyle y=t^2\).

Keep in mind, when taking the derivative with respect to \(\displaystyle x\), \(\displaystyle y\) is treated as a constant, and when taking the derivative with respect to \(\displaystyle y\), \(\displaystyle x\) is treated as a constant.

\(\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\)

\(\displaystyle \frac{dz}{dt}=(e^{5xy}*5y)(1/2t^{-1/2})+(e^{5xy}*5x)(2t)\)

To put \(\displaystyle \frac{dz}{dt}\) solely in terms of \(\displaystyle x\) and  \(\displaystyle y\), we substitute the definitions of \(\displaystyle x\) and \(\displaystyle y\)  given in the question,  \(\displaystyle x=\sqrt{t}\) and \(\displaystyle y=t^2\).

\(\displaystyle \frac{dz}{dt}=(e^{5t^{1/2}(t^2)}*5t^2)(1/2t^{-1/2})+(e^{5t^{1/2}(t^2)}*5t^{1/2})(2t)\)

\(\displaystyle \frac{dz}{dt}=(e^{5t^{5/2}}*5t^2)(1/2t^{-1/2})+(e^{5t^{5/2}}*5t^{1/2})(2t)\)

\(\displaystyle \frac{dz}{dt}=(e^{5t^{5/2}}*5/2t^{3/2})+(e^{5t^{5/2}}*10t^{3/2})\)

\(\displaystyle \frac{dz}{dt}=e^{5t^{5/2}}(5/2t^{3/2}+10t^{3/2})\)

\(\displaystyle \frac{dz}{dt}=\frac{25}{2}t^{3/2}e^{5t^{5/2}}\)

Find \(\displaystyle \frac{dz}{dt}\) if \(\displaystyle z=x^2+7xy^2\), \(\displaystyle x=1/t\) and \(\displaystyle y=5t^2-t\).

Keep in mind, when taking the derivative with respect to \(\displaystyle x\), \(\displaystyle y\) is treated as a constant, and when taking the derivative with respect to \(\displaystyle y\), \(\displaystyle x\) is treated as a constant.

\(\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\)

\(\displaystyle \frac{dz}{dt}=(2x+7y^2)(-1/t^2)+(14xy)(10t-1)\)

To put \(\displaystyle \frac{dz}{dt}\) solely in terms of \(\displaystyle x\) and  \(\displaystyle y\), we substitute the definitions of \(\displaystyle x\) and \(\displaystyle y\)  given in the question,  \(\displaystyle x=1/t\) and \(\displaystyle y=5t^2-t\).

\(\displaystyle \frac{dz}{dt}=[2(1/t)+7(5t^2-t)^2](-1/t^2)+[14(1/t)(5t^2-t)](10t-1)\)

\(\displaystyle \frac{dz}{dt}=[2/t+7(25t^4-10t^3+t^2)](-1/t^2)+[(14/t)(5t^2-t)](10t-1)\)

\(\displaystyle \frac{dz}{dt}=[2/t+175t^4-70t^3+7t^2](-1/t^2)+(70t-14)(10t-1)\)

\(\displaystyle \frac{dz}{dt}=-2/t^3-175t^2+70t-7+700t^2-140t-70t+14\)

\(\displaystyle \frac{dz}{dt}= -175t^2+700t^2 +70t-140t-70t +14-7-2/t^3\)

\(\displaystyle \frac{dz}{dt}= 525t^2-140t+7-2/t^3\)

Find \(\displaystyle \frac{dz}{dt}\) if \(\displaystyle z=sin(x)cos(y)\), \(\displaystyle x=t^4\) and \(\displaystyle y=7-t\).

Keep in mind, when taking the derivative with respect to \(\displaystyle x\), \(\displaystyle y\) is treated as a constant, and when taking the derivative with respect to \(\displaystyle y\), \(\displaystyle x\) is treated as a constant.

\(\displaystyle \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\)

\(\displaystyle \frac{dz}{dt}=[cos(y)cos(x)](4t^3)+[sin(x)(-sin(y))](-1)\)

\(\displaystyle \frac{dz}{dt}=4t^3cos(x)cos(y)+sin(x)sin(y)\) 

To put \(\displaystyle \frac{dz}{dt}\) solely in terms of \(\displaystyle x\) and  \(\displaystyle y\), we substitute the definitions of \(\displaystyle x\) and \(\displaystyle y\)  given in the question,  \(\displaystyle x=t^4\) and \(\displaystyle y=7-t\).

\(\displaystyle \frac{dz}{dt}=4t^3cos(t^4)cos(7-t)+sin(t^4)sin(7-t)\)

Compute \(\displaystyle \frac{df}{dt}\) for the function \(\displaystyle f(x,y,z)=3xe^{z+y}\) where the variables \(\displaystyle x, y\), and \(\displaystyle z\) are functions of the of the parameter \(\displaystyle t\):

\(\displaystyle x = t\: \: \:\: \: \: \:y=\ln(t)\: \: \: \: \: \: \: \:z=t^2\)

Because each function is given in terms of the parameter \(\displaystyle t\) we can conveniently write the function as a function of \(\displaystyle t\) alone: 

\(\displaystyle f(x(t), y(t), z(t))=f(t)\)

Solution 1

For the function described in this problem, the function can be written strictly in terms of the parameter \(\displaystyle t\) by simply substituting the definitions given for \(\displaystyle x\), \(\displaystyle y,\) or \(\displaystyle z\). 

\(\displaystyle f(x,y,z)=f(t, \: \ln t,\: \:t^2 ) )=3te^{t^2+\ln t}=3t^2e^{t^2}\)

 \(\displaystyle f(t)=3t^2e^{t^2}\)                                                         

So now we have the function written in the form of a single variable function of \(\displaystyle t.\) We can use the usual chain-rule. 

Apply the product rule and and the chain-rule:  

\(\displaystyle \frac{df}{dt}=3t^2\left(\frac{d}{dt}e^{t^2} \right )+e^{t^2}\left(\frac{d}{dt}3t^2 \right )=6te^{t^2}+3t^2e^{t^2}(2t)\)                                              

 \(\displaystyle =6te^{t^2}+6t^3e^{t^2}\)                                                        

 \(\displaystyle =6te^{t^2}(1+t^2)\)

Solution 2

The derivative of \(\displaystyle f\) with respect to \(\displaystyle t\) will be the sum of the derivatives of each variable \(\displaystyle x\), \(\displaystyle y\), and \(\displaystyle z\) with respect to \(\displaystyle t.\)

\(\displaystyle \frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}\)                                (1) 

In Equation (1) we use the partial derivative notation for a derivative of \(\displaystyle f\) with respect to the variables \(\displaystyle x\), \(\displaystyle y\), and \(\displaystyle z\) since it is a multi-variable function, we have to specify which variable we are differentiating unless we are differentiating with respect to \(\displaystyle t.\)

We return to the standard derivative notation when computing the derivatives with respect to \(\displaystyle t\) since each function \(\displaystyle f\), \(\displaystyle x\), \(\displaystyle y\), and \(\displaystyle z\) can be written as a single variable function of \(\displaystyle t.\) Now simply apply (3) to the function term-by-term: 

\(\displaystyle \frac{\partial f}{\partial x}\frac{dx}{dt}=3e^{z+y}\)

\(\displaystyle \frac{\partial f}{\partial y}\frac{dy}{dt}=3xe^{z+y}\frac{1}{t}\)

\(\displaystyle \frac{\partial f}{\partial z}\frac{dz}{dt}=3xe^{z+y}(2t)=6xte^{z+y}\)

Now ad the terms to get an expression for \(\displaystyle \frac{df}{dt}\) and then rewrite in terms of \(\displaystyle t.\)

 \(\displaystyle \frac{df}{dt}=3xe^{z+y}\frac{1}{t}+6xte^{z+y}+3e^{z+y}\)

\(\displaystyle =3e^{z+y}(1+\frac{x}{t}+2xt)\)

  \(\displaystyle =3e^{t^2+\ln t}\left(1+\frac{t}{t}+2t^2\right)\)

 \(\displaystyle =6e^{t^2+\ln t}\left(1+t^2\right)\)

\(\displaystyle =6te^{t^2}\left(1+t^2\right)\)

Therefore: 

\(\displaystyle \frac{df}{dt}=6te^{t^2}\left(1+t^2\right)\)

 \(\displaystyle \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x} \right )=2\)

\(\displaystyle \frac{\partial f}{\partial x}(0,1)=1\) 

\(\displaystyle f(1,0)=2\)

The first step is to integrate the function with respect to \(\displaystyle y\) since that is the outer most derivative in this problem. 

\(\displaystyle \frac{\partial f}{\partial x}= \int \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x} \right )dy=\int 2dy=2y+C_1\)

\(\displaystyle f(x,y)=\int \frac{\partial f}{\partial x}dx=\int (2y+C_1)dx=(2y+C_1)x+C_2\)

Now apply the constraints to compute the two constants of integration. 

 \(\displaystyle \frac{\partial f}{\partial x}(0,1)=2+C_1 = 1 \: \: \: \: \: \: \:\Rightarrow \: \: \: \: \: \: \:C_1=-1\)

So far we have: 

\(\displaystyle f(x,y)=(2y-1)x+C_2\)

Now apply the second constraint given: 

\(\displaystyle f(1,0)=2\: \: \: \: \: \:\ \Rightarrow \: \: \: \: \: -1+C_2 =2\: \: \: \: \: \Rightarrow \: \: \: \:C_2 =3\)

\(\displaystyle f(x,y)=(2y-1)x+3\)

The question is really asking for a tangent plane, so lets first find partial derivatives and then plug in the point.

\(\displaystyle f(x,y)=6+\frac{x^2}{4}+\frac{y^2}{16}\), \(\displaystyle f(-2,4)=6+\frac{(-2)^2}{4}+\frac{(4)^2}{16}=6+1+1=8\)

\(\displaystyle f_x(x,y)=\frac{x}{2}\), \(\displaystyle f_x(-2,4)=\frac{-2}{2}=-1\)

\(\displaystyle f_y(x,y)=\frac{y}{8}\), \(\displaystyle f_y(-2,4)=\frac{4}{8}=\frac{1}{2}\)

Remember that we need to build the linear approximation general equation which is as follows.

\(\displaystyle L(x,y)\approx f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\)

\(\displaystyle L(x,y)\approx 8-(x+2)+\frac{1}{2}(y-4)\)

\(\displaystyle L(x,y)\approx 8-x-2+\frac{1}{2}y-2\)

\(\displaystyle L(x,y)\approx 4-x+\frac{1}{2}y\)

To find the equation of the tangent plane, we use the formula

\(\displaystyle z- z_0 = f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\).

Taking partial derivatives, we have

\(\displaystyle f_x(x,y) =\) \(\displaystyle 2y-e^{-x}\)

\(\displaystyle f_y(x,y) = 2x\)

Substituting our \(\displaystyle x, y\) values into these, we get

\(\displaystyle f_x(0,1) =1\)

\(\displaystyle f_y(0,1) = 0\)

Substituting our point into \(\displaystyle x_0, y_0, z_0\), and partial derivative values in the formula we get

\(\displaystyle z - 1 = 1(x-0)+0(y-1)\)

\(\displaystyle z - 1 = x\)

\(\displaystyle -x+z =1\).

We are just asking for the equation of the tangent plane:

Step 1: Find \(\displaystyle f(x,y)\)

\(\displaystyle f(x,y)=2+\frac {(-3)^2}{9}+\frac {2^2}{4}=2+\frac {9}{9}+\frac {4}{4}=2+1+1=4\)

Step 2: Take the partial derivative of \(\displaystyle \frac {x^2}{9}\) with respect with (x,y):

\(\displaystyle \partial(\frac {x^2}{9})=\frac {2x}{9}\)

Step 3: Evaluate the partial derivative of x at \(\displaystyle -3\)

\(\displaystyle \frac {2(3)}{9}=\frac {6}{9}=\frac {2}{3}\)

Step 4: Take the partial derivative of \(\displaystyle \frac {y^2}{4}\) with respect to \(\displaystyle (x,y)\):

\(\displaystyle \partial (\frac{y^2}{4})=\frac {2y}{4}\)

Step 5: Evaluate the partial derivative at \(\displaystyle y=2\)

\(\displaystyle \frac {2(2)}{4}=\frac {4}{4}=1\).

Step 6: Convert (x,y) back into binomials:

\(\displaystyle x\rightarrow (x+3)\)
\(\displaystyle y\rightarrow (y-2)\)

Step 7: Write the equation of the tangent line:

\(\displaystyle 4+\frac {2}{3}(x+3)+(y-2)\)