The first derivative is easy:

f'(x) = 12x³ – 15x² – 4

The slope of the tangent line is found by calculating f'(40) = 12 * 40³ – 15 * 40² – 4 = 768,000 – 24,000 – 4 = 743,996

To find the slope of a tangent line, we need to find the first derivative of the function at that point. In other words, we need y'(6).

\(\displaystyle y(b)=b^7-b^5+4b^2\)

Taking the first derivative using the Power Rule \(\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}\) we get the following.

\(\displaystyle y'(b)=7b^6-5b^4+8b\)

Substituting in 6 for b and solving we get:

\(\displaystyle y'(6)=7\cdot6^6-5\cdot6^4+8\cdot6=320160\).

So our answer is 320160

To find the slope of a tangent line, we need the first derivative.

Recall that to find the first derivative of a polynomial, we need to decrease each exponent by one and multiply by the original number.

\(\displaystyle j(x)=\frac{4x^5}{5}-x^4+x^2\)

\(\displaystyle j'(x)=4x^4-4x^3+2x\)

The slope of the tangent line can be found easily via derivatives. To find the slope of the tangent line at s=16, find b'(16) using the power rule on each term which states:

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

Applying this rule we get:

\(\displaystyle b(s)=14s^2+6s-4\)

\(\displaystyle b'(s)=28s+6\)

\(\displaystyle b'(16)=28\cdot 16+6=454\)

Therefore, the slope we are looking for is 454.

To find the slope of the line at that point, find the derivative of f(x) and plug in that point. 

\(\displaystyle f'(x)= \frac{2x}{x^2 +2}+17sin(17x)\)

Remember that the derivative of \(\displaystyle cos(bx) = -bsin(bx)\) and the derivative of   \(\displaystyle ln(u) = \frac{u'}{u}\)

Now plug in \(\displaystyle x=\) \(\displaystyle \pi\)

\(\displaystyle f'(\pi )=\frac{2\pi}{\pi ^2 +2}=.529\)

The first step here is to integrate \(\displaystyle f''(x)\) in order to get \(\displaystyle f'(x)\).

\(\displaystyle \int f''(x)dx =\int(x+2)dx=\frac{x^2}{2} +2x +C\)

Here the problem tells us that the integration constant \(\displaystyle C=0\), so

\(\displaystyle f'(x)=\frac{x^2}{2} + 2x\)

Plug in \(\displaystyle x=2\) here

\(\displaystyle f'(2)=6\)

The slope of a curve at any point is equal to the derivative of the curve at that point.

Remembering that the derivative of \(\displaystyle ln(x)\rightarrow \frac{1}{x}\) and using the power rule on the second term we find the derivative to be:

\(\displaystyle \frac{dy}{dx}=\frac{1}{x}+6x\).

Pluggin in \(\displaystyle x=1.5\) we find that the slope is \(\displaystyle \frac{29}{3}\).

Find the line tangent to \(\displaystyle y=e^{5x}\) at \(\displaystyle x=3\).

First, we find \(\displaystyle y\):

\(\displaystyle y=e^{5x}=e^{5\cdot3}=e^{15}\)

Next, we find the derivative:

\(\displaystyle \frac{dy}{dx}=5e^{5x}\)

Therefore, the slope at \(\displaystyle x=3\) is:

\(\displaystyle \frac{dy}{dx}=5e^{5x}=5e^{5\cdot3}=5e^{15}\).

Using point-slope form, we can write the tangent line:

\(\displaystyle y=5e^{15}(x-3)+e^{15}\)

Simplifying this gives us:

\(\displaystyle y=5e^{15}x-14e^{15}\)

The other point of the triangle must be at \(\displaystyle (0,6)\) as it must be equidistant from the other two points of the triangle. Since all points on the y-axis are \(\displaystyle 6\) units away from the other points in the \(\displaystyle x\) direction, the third point must be equidistant in the \(\displaystyle y\) direction from both \(\displaystyle (6,2)\) and \(\displaystyle (6,10)\). The distance between these points is \(\displaystyle 8\), so the third point must have a y-value of \(\displaystyle 2+(8/2)=6\). The third point is now at \(\displaystyle (0,6)\) so the slope of the line from \(\displaystyle (0,6)\) to \(\displaystyle (6,10)\) is as follows.

\(\displaystyle \\m=\frac{y_2-y_1}{x_2-x_1}\\ \\=\frac{10-6}{6-0}\\ \\ =\frac{4}{6}=\frac{2}{3}\)

We must take the derivative of the function using the chain rule yielding \(\displaystyle f'(x)=2\cos(2x)\).

The chain rule is \(\displaystyle [f(g(x))]'=f'(g(x))\cdot g'(x)\).

Also remember that the derivative of \(\displaystyle sin(x)\) is \(\displaystyle cos(x)\).

Applying these rules we get the following.

\(\displaystyle \\f(x)=\sin(2x)\\ \\ u=2x\\ \\ \frac{du}{dx}=2\\ \\ \frac{d}{dx}\sin(u)=\cos(u)\frac{du}{dx}=2\cos(u)\\ \\ f'(x)=2\cos(2x)\)

Plugging in the value for \(\displaystyle x\) we get \(\displaystyle 2\cos(\pi)\) which is \(\displaystyle -2\).