The first derivative of the function is equal to

\(\displaystyle f'(x)=(2x\sin(x)+x^2\cos(x))(e^{x^2\sin(x)})\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)\)

Finally, to evaluate the derivative at x=0, simply plug this into the first derivative function:

\(\displaystyle f'(0)=(0+0)e^{0}=0\)

The first derivative of the function is equal to

\(\displaystyle f'(x)=-2\sin^2(2x)+2\cos^2(2x)\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

The second derivative - found by taking the derivative of the first derivative function - is equal to

\(\displaystyle f''(x)=-16\sin(2x)\cos(2x)\)

and was found using the same rules as above.

Use the power rule to find the derivative.

\(\displaystyle \frac{d}{dy}y^2=2y\)

\(\displaystyle \frac{d}{dy}5y=5\)

\(\displaystyle \frac{d}{dy}-2=0\)

Thus, the derivative is \(\displaystyle 2y+5\).

If \(\displaystyle f(x)=C (constant))\), then the derivative is \(\displaystyle f'(x) = 0\).

If \(\displaystyle f(x)= x^C\), then the derivative is \(\displaystyle f'(x) = Cx^{C-1}\).

If \(\displaystyle f(x)= ln(x)\), then the derivative is  \(\displaystyle f'(x)= \frac{1}{x}\).

If \(\displaystyle f(x)= e^x\), the the derivative is \(\displaystyle f'(x)=e^x\).

If \(\displaystyle f(x)= \sin x\), then the derivative is \(\displaystyle f'(x)= \cos x\).

There are many other rules for the derivatives for trig functions. 

If \(\displaystyle F(x)= (f*g)(x)\), then the derivative is \(\displaystyle F'(x)= f'(g(x))g'(x)\). This is known as the chain rule.

In this case, we must find the derivative of the following: \(\displaystyle f(x)= (9x^2 - 2x + 5)^4\)

That is done by doing the following: \(\displaystyle f'(x)= (4)(9x^2 - 2x + 5)^{4-1} ((2)9x^{2-1} - (1)2x^{1-1} + 0)\)

Therefore, the answer is: \(\displaystyle f'(x)= 4(9x^2 - 2x + 5)^{3} (18x - 2)\)

If \(\displaystyle f(x)=C (constant))\), then the derivative is \(\displaystyle f'(x) = 0\).

If \(\displaystyle f(x)= x^C\), then the derivative is \(\displaystyle f'(x) = Cx^{C-1}\).

If \(\displaystyle f(x)= ln(x)\), then the derivative is  \(\displaystyle f'(x)= \frac{1}{x}\).

If \(\displaystyle f(x)= e^x\), the the derivative is \(\displaystyle f'(x)=e^x\).

If \(\displaystyle f(x)= \sin x\), then the derivative is \(\displaystyle f'(x)= \cos x\).

There are many other rules for the derivatives for trig functions. 

If \(\displaystyle F(x)= (f*g)(x)\), then the derivative is \(\displaystyle F'(x)= f'(g(x))g'(x)\). This is known as the chain rule.

In this case, we must find the derivative of the following: \(\displaystyle f(x)= 23 + e\)

That is done by doing the following: \(\displaystyle f'(x)= 0 + 0\)

Therefore, the answer is: \(\displaystyle f'(x)= 0\)

If \(\displaystyle f(x)=C (constant))\), then the derivative is \(\displaystyle f'(x) = 0\).

If \(\displaystyle f(x)= x^C\), then the derivative is \(\displaystyle f'(x) = Cx^{C-1}\).

If \(\displaystyle f(x)= ln(x)\), then the derivative is  \(\displaystyle f'(x)= \frac{1}{x}\).

If \(\displaystyle f(x)= e^x\), the the derivative is \(\displaystyle f'(x)=e^x\).

If \(\displaystyle f(x)= \sin x\), then the derivative is \(\displaystyle f'(x)= \cos x\).

There are many other rules for the derivatives for trig functions. 

If \(\displaystyle F(x)= (f*g)(x)\), then the derivative is \(\displaystyle F'(x)= f'(g(x))g'(x)\). This is known as the chain rule.

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

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

Therefore, the answer is: \(\displaystyle f'(x)= \frac{1}{x} + \frac{1}{x}\)

If \(\displaystyle f(x)=C (constant))\), then the derivative is \(\displaystyle f'(x) = 0\).

If \(\displaystyle f(x)= x^C\), then the derivative is \(\displaystyle f'(x) = Cx^{C-1}\).

If \(\displaystyle f(x)= ln(x)\), then the derivative is  \(\displaystyle f'(x)= \frac{1}{x}\).

If \(\displaystyle f(x)= e^x\), the the derivative is \(\displaystyle f'(x)=e^x\).

If \(\displaystyle f(x)= \sin x\), then the derivative is \(\displaystyle f'(x)= \cos x\).

There are many other rules for the derivatives for trig functions. 

If \(\displaystyle F(x)= (f*g)(x)\), then the derivative is \(\displaystyle F'(x)= f'(g(x))g'(x)\). This is known as the chain rule.

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

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

Therefore, the answer is: \(\displaystyle f'(x)= \cos(4x^4 - x) (16x^{3} -1)\)

If \(\displaystyle f(x)=C (constant))\), then the derivative is \(\displaystyle f'(x) = 0\).

If \(\displaystyle f(x)= x^C\), then the derivative is \(\displaystyle f'(x) = Cx^{C-1}\).

If \(\displaystyle f(x)= ln(x)\), then the derivative is  \(\displaystyle f'(x)= \frac{1}{x}\).

If \(\displaystyle f(x)= e^x\), the the derivative is \(\displaystyle f'(x)=e^x\).

If \(\displaystyle f(x)= \sin x\), then the derivative is \(\displaystyle f'(x)= \cos x\).

There are many other rules for the derivatives for trig functions. 

If \(\displaystyle F(x)= (f*g)(x)\), then the derivative is \(\displaystyle F'(x)= f'(g(x))g'(x)\). This is known as the chain rule.

In this case, we must find the derivative of the following: \(\displaystyle f(x)= e^{3x^6 - e^x}\)

That is done by doing the following: \(\displaystyle f'(x)= e^{3x^6 - e^x} ((6)3x^{6-1} - e^x)\)

Therefore, the answer is: \(\displaystyle f'(x)= e^{3x^6 - e^x} (18x^{5} - e^x)\)

If \(\displaystyle f(x)=C (constant))\), then the derivative is \(\displaystyle f'(x) = 0\).

If \(\displaystyle f(x)= x^C\), then the derivative is \(\displaystyle f'(x) = Cx^{C-1}\).

If \(\displaystyle f(x)= ln(x)\), then the derivative is  \(\displaystyle f'(x)= \frac{1}{x}\).

If \(\displaystyle f(x)= e^x\), the the derivative is \(\displaystyle f'(x)=e^x\).

If \(\displaystyle f(x)= \sin x\), then the derivative is \(\displaystyle f'(x)= \cos x\).

There are many other rules for the derivatives for trig functions. 

If \(\displaystyle F(x)= (f*g)(x)\), then the derivative is \(\displaystyle F'(x)= f'(g(x))g'(x)\). This is known as the chain rule.

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

This simplifies to: \(\displaystyle f(x)= e^{ln(x)} =x\)

That is done by doing the following: \(\displaystyle f'(x)= 1\)

Therefore, the answer is: \(\displaystyle f'(x)= 1\)

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

Taking the derivative of the function

\(\displaystyle f(x)=x^3+19x+2\)

The slope of the tangent is

\(\displaystyle f'(x)=3x^2+19\)

\(\displaystyle f'(2)=3(2)^2+19=31\)