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

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.

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

$\displaystyle \frac{dz}{dt}=(ycos(xy))(-5t^4+4)+(xcos(xy))(2)$ 

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,  $\displaystyle x=-t^5+4t$ and $\displaystyle y=2t-3$.

$\displaystyle \frac{dz}{dt}=\left [ (2t-3)cos\left [ (-t^5+4t)(2t-3) \right ] \right ](-5t^4+4) +\left [ (-t^5+4t)cos\left [ (-t^5+4t)(2t-3) \right ] \right ](2)$

$\displaystyle \frac{dz}{dt}=\left [ (2t-3)(-5t^4+4)cos\left [ (-2t^6+8t^2+3t^5-12t) \right ] \right ] +\left [ 2(-t^5+4t)cos\left [ (-2t^6+8t^2+3t^5-12t) \right ] \right ]$

$\displaystyle \frac{dz}{dt}=(-10t^5+8t+15t^4-12)cos(-2t^6+8t^2+3t^5-12t) +(-2t^5+8t)cos(-2t^6+8t^2+3t^5-12t)$$\displaystyle \frac{dz}{dt}=(-10t^5+8t+15t^4-12-2t^5+8t)cos(-2t^6+8t^2+3t^5-12t)$

$\displaystyle =(-12t^5+15t^4+16t-12)cos(-2t^6+8t^2+3t^5-12t)$

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: 

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

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

In order to find $\displaystyle \frac{\partial f}{dz}$, we need to take the derivative of $\displaystyle f$ in respect to $\displaystyle z$, and treat $\displaystyle x$, and $\displaystyle y$ as constants. We also need to remember what the derivatives of natural log, exponential functions and power functions are for single variables.

Natural Log:

$\displaystyle f(x)=\ln(g(x))$

$\displaystyle f'(x)=\frac{g'(x)}{g(x)}$

Exponential Functions:

$\displaystyle f(x)=e^{g(x)}$

$\displaystyle f'(x)=g'(x)e^{g(x)}$

Power Functions:

$\displaystyle f(x)=x^n$

$\displaystyle f'(x)=nx^{n-1}$

$\displaystyle \frac{\partial f}{dz}=10xyz^9-xy(\ln(yz))^{-2}(\frac{y}{yz})+10e^z+0$

$\displaystyle \frac{\partial f}{dz}=10xyz^9-\frac{xy}{z\ln^2(yz)}+10e^z$

In order to find $\displaystyle \frac{\partial f}{dy}$, we need to take the derivative of $\displaystyle f$ in respect to  $\displaystyle y$, and treat $\displaystyle x$, and $\displaystyle z$ as constants. We also need to remember what the derivatives of natural log, exponential functions and power functions are for single variables.

Natural Log:

$\displaystyle f(x)=\ln(g(x))$

$\displaystyle f'(x)=\frac{g'(x)}{g(x)}$

Exponential Functions:

$\displaystyle f(x)=e^{g(x)}$

$\displaystyle f'(x)=g'(x)e^{g(x)}$

Power Functions:

$\displaystyle f(x)=x^n$

$\displaystyle f'(x)=nx^{n-1}$

$\displaystyle \frac{\partial f}{dy}=xz^{10}+x(\ln(yz))^{-1}-xy(\ln(yz))^{-2}\cdot\frac{z}{yz}+0+xe^{xy}$

$\displaystyle \frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$