By deriving the trigonometric function \(\displaystyle \sin(x)\) , we are given \(\displaystyle x'\cos(x)\).  The \(\displaystyle x'\) is usually ignored unless \(\displaystyle x\) is being used to express a different function, where it's own derivative is not \(\displaystyle 1\).  In the case of \(\displaystyle \sin(3x^2+7)\), we take the derivative of not only the \(\displaystyle \sin\) function but also the \(\displaystyle 3x^2+7\).  We then multiply the two derivatives and end up with our answer, \(\displaystyle \left 6x[\cos(3^2+7)]\).

To consider finding the slope, let's discuss the topic of the gradient.

For a function \(\displaystyle f(x_1,x_2,...,x_n)\), the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rule will be necessary:

Derivative of an exponential: \(\displaystyle d[e^u]=e^udu\)

Note that u and v may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Evaluating the derivatives of \(\displaystyle f(xy)=\frac{1}{e^{xy}}=e^{-xy}\) at the point \(\displaystyle (1,3)\)

x:

\(\displaystyle \frac{\delta f}{\delta x}=-ye^{-xy};\frac{\delta f}{\delta x}=-3e^{-(-1)3}=-3e^3\)

y:

\(\displaystyle \frac{\delta f}{\delta y}=-xe^{-xy};\frac{\delta f}{\delta y}=-(-1)e^{-(-1)(3)}=e^3\)

Thus the slope is 

\(\displaystyle -3e^3\widehat{i}+e^3\widehat{j}\)

To consider finding the slope, let's discuss the topic of the gradient.

For a function \(\displaystyle f(x_1,x_2,...,x_n)\), the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

It is essentially the slope of a multi-dimensional function at any given point

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Taking the partials of \(\displaystyle f(x,y)=x^2y+\frac{x}{y^2}=x^2y+xy^{-2}\) at the point \(\displaystyle (3,1)\) 

x:

\(\displaystyle \frac{\delta f}{\delta x}=2xy+\frac{1}{y^2}=2(3)(1)+\frac{1}{1^2}=7\)

y:

\(\displaystyle \frac{\delta f}{\delta y}=x^2-2\frac{x}{y^3}=3^2-2\frac{3}{1^3}=3\)

Thus the slope is \(\displaystyle 7\widehat{i}+3\widehat{j}\)

To consider finding the slope, let's discuss the topic of the gradient.

For a function \(\displaystyle f(x_1,x_2,...,x_n)\), the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

It is essentially the slope of a multi-dimensional function at any given point.

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Taking the partials of \(\displaystyle f(x,y,z) = \frac{x}{y+z}=x(y+z)^{-1}\) at the point \(\displaystyle (-1,0,2)\) 

x:

\(\displaystyle \frac{\delta f}{\delta x}=\frac{1}{y+z}=\frac{1}{0+2}=0.5\)

y:

\(\displaystyle \frac{\delta f}{\delta y}=\frac{-x}{(y+z)^2}=\frac{-(-1)}{(0+2)^2}=0.25\)

z:

\(\displaystyle \frac{\delta f}{\delta z}=\frac{-x}{(y+z)^2}=\frac{-(-1)}{(0+2)^2}=0.25\)

The slope is

\(\displaystyle 0.5\widehat{i}+0.25\widehat{j}+0.25\widehat{k}\)

We want to find the equation of the normal line to \(\displaystyle f(x)\) at \(\displaystyle x=10\). The reason for this is that if the fruit goes down a straight tunnel at a right angle to the tangent at that point, its trajectory is uniquely given by the perpendicular line at that point. That is the very definition of a normal line. The slope of the tangent line, of course, is given by the derivative:

\(\displaystyle f(x)= \frac{1}{32}x^{2}\)

\(\displaystyle f'(x)= \frac{2}{32}x\)    so...

\(\displaystyle f'(10)= \frac{20}{32} = \frac{5}{8}\)

We know the slope of a perpendicular line has negative reciprocal slope of the tangent, so the slope of the normal line is :

\(\displaystyle \frac{-1}{\frac{5}{8}}= -\frac{8}{5}\)

So, using the point slope formula, the formula for the normal line is:

\(\displaystyle y- \frac{25}{8} = -\frac{8}{5}(x-10)\)

\(\displaystyle y= 16-\frac{8}{5}x +\frac{25}{8} = \frac{153}{8}-\frac{8}{5}x\)

Now we use this to find the end of the tunnel. Since the x-coordinate of the tunnels length is five, we know its exit is at \(\displaystyle x=15\), so the fruit exits the tunnel at:

\(\displaystyle \frac{153}{8}-\frac{8}{5}(15) = -\frac{39}{8}\)

To consider finding the slope, let's discuss the topic of the gradient.

For a function \(\displaystyle f(x_1,x_2,...,x_n)\), the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative: \(\displaystyle d[cos(u)]=-sin(u)du\)

Product rule: \(\displaystyle d[uv]=udv+vdu\)

Note that u and v may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Taking the partials of \(\displaystyle f(x,y)=x^2ycos(y)\) at the point \(\displaystyle (5,2\pi)\)

x:

\(\displaystyle \frac{\delta f}{\delta x}=2xycos(y)=2(5)(2\pi)cos(2\pi)=20\pi\)

y:

\(\displaystyle \frac{\delta f}{\delta y}=x^2cos(y)-x^2ysin(y)=5^2cos(2\pi)-5^2(2\pi)sin(2\pi)=25\)

The slope is 

\(\displaystyle 20\pi\widehat{i}+25\widehat{j}\)

To consider finding the slope, let's discuss the topic of the gradient.

For a function \(\displaystyle f(x_1,x_2,...,x_n)\), the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rule will be necessary:

Trigonometric derivative: \(\displaystyle d[sin(u)]=cos(u)du\)

Note that u and v may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Taking the derivatives of \(\displaystyle f(x,y)=ysin(2x)+2xsin(y)\) at the point \(\displaystyle (\pi,\pi)\)

x:

\(\displaystyle \frac{\delta f}{\delta x}=2ycos(2x)+2sin(y)=2\pi cos(2\pi)+2sin(\pi)=2\pi\)

y:

\(\displaystyle \frac{\delta f}{\delta y}=sin(2x)+2xcos(y)=sin(2\pi)+2\pi cos(\pi)=-2\pi\)

The slope is

\(\displaystyle 2\pi\widehat{i}-2\pi\widehat{j}\)

To consider finding the slope, let's discuss the topic of the gradient.

For a function \(\displaystyle f(x_1,x_2,...,x_n)\), the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

Derivative of an exponential: \(\displaystyle d[e^u]=e^udu\)

Derivative of a natural log: \(\displaystyle d[ln(u)]=\frac{du}{u}\)

Trigonometric derivative: \(\displaystyle d[sin(u)]=cos(u)du\)

Product rule: \(\displaystyle d[uv]=udv+vdu\)

Note that u and v may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Taking the partial derivatives of \(\displaystyle f(x,y)=e^xln(y)sin(xy)\) at the point \(\displaystyle (3,3)\)

x:

\(\displaystyle \frac{\delta f}{\delta x}=e^xln(y)sin(xy)+ye^xln(y)cos(xy)\)

\(\displaystyle \frac{\delta f}{\delta x}=e^3ln(3)sin(3(3))+(3)e^3ln(3)cos(3(3))=-51.2\)

y:

\(\displaystyle \frac{\delta f}{\delta y}=\frac{e^xsin(xy)}{y}+xe^xln(y)cos(xy)\)

\(\displaystyle \frac{\delta f}{\delta y}=\frac{e^3sin(3(3))}{3}+3e^3ln(3)cos(3(3))=-57.6\)

The slope is

\(\displaystyle -51.2\widehat{i}-57.6\widehat{j}\)

To consider finding the slope, let's discuss the topic of the gradient.

For a function \(\displaystyle f(x_1,x_2,...,x_n)\), the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rule will be necessary:

Derivative of an exponential: \(\displaystyle d[e^u]=e^udu\)

Note that u and v may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Take the partial derivatives of \(\displaystyle f(x,y,z)=x^2e^z+ye^{3z}\) at the point \(\displaystyle (3,4, ln(2))\)

x:

\(\displaystyle \frac{\delta f}{\delta x}=2xe^z=2(3)e^{ln(2)}=12\)

y:

\(\displaystyle \frac{\delta f}{\delta y}=e^{3z}=e^{3ln(2)}=8\)

z:

\(\displaystyle \frac{\delta f}{\delta z}=x^2e^z+3ye^{3z}=3^2e^{ln(2)}+3(4)e^{3ln(2)}=18+96=114\)

The slope is

\(\displaystyle 12\widehat{i}+8\widehat{j}+114\widehat{k}\)

To consider finding the slope, let's discuss the topic of the gradient.

For a function \(\displaystyle f(x_1,x_2,...,x_n)\), the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

Derivative of an exponential: \(\displaystyle d[e^u]=e^udu\)

Trigonometric derivative: \(\displaystyle d[sin(u)]=cos(u)du\)

Note that u and v may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Take the partial derivatives of \(\displaystyle f(x,y)=sin(xe^y)\) at the point \(\displaystyle (2,3)\)

x:

\(\displaystyle \frac{\delta f}{\delta x}=e^ycos(xe^y)=e^3cos(2e^3)=-15.7\)

y:

\(\displaystyle \frac{\delta f}{\delta y}=xe^ycos(xe^y)=2e^3cos(2e^3)=-31.5\)

The slope is \(\displaystyle -15.7\widehat{i}-31.5\widehat{j}\)