In order to find the slope of a function, you must first differentiate that function.

In this case, the derivative of the given function is: \(\displaystyle f'(x)= 2e^{2x}\)

Then, plug \(\displaystyle x=1\) into the function to get the slope: \(\displaystyle f'(1)= 2e^{2(1)}\)

Therefore, the slope is: \(\displaystyle 14.8\)

In order to find the slope of a function, you must first differentiate that function.

In this case, the derivative of the given function is: \(\displaystyle f'(x)= -\sin(x^3) 3x^2\)

Then, plug \(\displaystyle x=3\) into the function to get the slope: \(\displaystyle f'(3)= -\sin((3)^3) 3(3)^2\)

Therefore, the slope is: \(\displaystyle -25.8\)

In order to find the slope of a function, you must first differentiate that function.

In this case, the derivative of the given function is: \(\displaystyle f'(x)= 15x^2 + 2(2x - ln(x))(2-\frac{1}{x})\)

Then, plug \(\displaystyle x=0\) into the function to get the slope: \(\displaystyle f'(0)= 15(0)^2 + 2(2(0) - ln(0))(2-\frac{1}{(0)})\)

Therefore, the slope is: \(\displaystyle undefined\)

In order to find the slope of a function, you must first differentiate that function.

In this case, the derivative of the given function is: \(\displaystyle f'(x)= \frac{1}{x} -\sin (\sin x) (\cos x)\)

Then, plug \(\displaystyle x=-1\) into the function to get the slope: \(\displaystyle f'(-1)= \frac{1}{(-1)} -\sin (\sin (-1)) (\cos (-1))\)

Therefore, the slope is: \(\displaystyle -0.6\)

Begin by writing the equations for a cube's dimensions. Namely its volume and diagonal in terms of the length of its sides:

\(\displaystyle V=s^3\)

\(\displaystyle d=s\sqrt{3}\)

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

\(\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}\)

\(\displaystyle \frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}\)

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and diagonal:

\(\displaystyle 3s^2\frac{ds}{dt}=(\phi)\sqrt{3}\frac{ds}{dt}\)

\(\displaystyle \phi=s^2\sqrt{3}\)

\(\displaystyle \phi =(13\sqrt{3})^2\sqrt{3}=507\sqrt{3}\)

In order to find the slope of a function, you must find its derivative. 

In this case, we must find the derivative of the following: \(\displaystyle f(x) = 12x^{\frac{1}{2}} + (6x^2 - 5x +2)^2\)

That is done by doing the following: \(\displaystyle f'(x) = (\frac{1}{2})12x^{(\frac{1}{2}-1)} + (2)(6x^2 - 5x +2)^{(2-1)}((2)6x^{(2-1)}- (1)5x^{(1-1)}+0 )\)

So, the derivative is: \(\displaystyle f'(x) = 6x^{-\frac{1}{2}} + 2(6x^2 - 5x +2) (12x- 5 )\)

Then, you plug \(\displaystyle x=3\) into the derivative: \(\displaystyle f'(3) = 6(3)^{-\frac{1}{2}} + 2(6(3)^2 - 5(3) +2) (12(3)- 5 )\)

Therefore, the answer is: \(\displaystyle 2,545.5\)

In order to find the slope of a function, you must find its derivative. 

In this case, we must find the derivative of the following: \(\displaystyle f(x) = \sin(x^2+2x-3)\)

That is done by doing the following: \(\displaystyle f'(x) = \cos (x^2+2x-3) ((2)x^{(2-1)} + (1)2x^{(1-1)} + 0)\)

Therefore, the derivative is: \(\displaystyle f'(x) = \cos (x^2+2x-3) (2x + 2)\)

Then, you plug \(\displaystyle x=0\) into the derivative: \(\displaystyle f'(0) = \cos ((0)^2+2(0)-3) (2(0) + 2)\)

Therefore, the answer is: \(\displaystyle -1.98\)

In order to find the slope of a function, you must find its derivative. 

In this case, we must find the derivative of the following: \(\displaystyle f(x) = ln(2x^3 + e^x)\)

That is done by doing the following: \(\displaystyle f'(x) = \frac{1}{2x^3+e^x} ((3)2x^{(3-1)} + e^x)\)

Therefore, the derivative is: \(\displaystyle f'(x) = \frac{6x^2+e^x}{2x^3 + e^x}\)

Then, you plug \(\displaystyle x=0\) into the derivative: \(\displaystyle f'(0) = \frac{6(0)^2+e^(0)}{2(0)^3 + e^(0)}\)

Therefore, the answer is: \(\displaystyle 1\)

In order to find the slope of a function, you must find its derivative. 

In this case, we must find the derivative of the following: \(\displaystyle f(x) = e^{4x}\)

That is done by doing the following: \(\displaystyle f'(x) = e^{4x} ((1)4x^{(1-1)})\)

Therefore, the derivative is: \(\displaystyle f'(x) = 4e^{4x}\)

Then, you plug \(\displaystyle x=1\) into the derivative: \(\displaystyle f'(1) = 4e^{4(1)}\)

Therefore, the answer is: \(\displaystyle 218.4\)

In order to find the slope of a function, you must find its derivative. 

In this case, we must find the derivative of the following: \(\displaystyle f(x) = \sin (ln(2x))\)

That is done by doing the following: \(\displaystyle f'(x) = \cos (ln(2x)) (\frac{1}{2x}) ((1)2x^{(1-1)})\)

Therefore, the derivative is: \(\displaystyle f'(x) = \frac{\cos (ln(2x))}{x}\)

Then, you plug \(\displaystyle x=1\) into the derivative: \(\displaystyle f'(1) = \frac{\cos (ln(2(1)))}{(1)}\)

Therefore, the answer is: \(\displaystyle 0.77\)