Math - Calculus I — Derivatives

What is the first derivative of $4x^2+\frac{2}{3}x$ ?

$8x+\frac{2}{3}$

$8\tfrac{2}{3}x^2$

Explanation

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{2-1})+(1*\frac{2}{3}x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{1})+(1*\frac{2}{3}x^{0})$

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(8x)+(1*\frac{2}{3}(1))$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=8x+\frac{2}{3}$

What is the first derivative of $5x^4+\sin x$ ?

$20x^{3}+\cos x$

$20x^{3}-\cos x$

$20x^3+4\cos x$

$5x^4\cos x +20x^3\sin x$

$20x^3+\frac{1}{\sin x}$

Explanation

Since we're adding terms, we take the derivative of each part separately. For , we can use the power rule, which states that we multiply the variable by the current exponent and then lower the exponent by one. For sine, we use our trigonometric derivative rules.

Remember, $\frac{\mathrm{d} }{\mathrm{d} x} \sin x=\cos x$ .

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(4*5x^{4-1})+\cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(20x^{3})+\cos x$

$\frac{\mathrm{d} }{\mathrm{d} x} \sin(2x)=?$

$2\cos(2x)$

$\cos(2x)$

$-\frac{1}{2}\cos(2x)$

$2\cos(x)$

$2\sin(2x)$

Explanation

For this problem we need to use the chain rule:

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(2x)=\frac{\mathrm{d} }{\mathrm{d} x} \sin(2x)*\frac{\mathrm{d} }{\mathrm{d} x}2x$

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(2x)=\cos(2x)*2$

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(2x)=2\cos(2x)$

What is the first derivative of $4x^2+\frac{2}{3}x$ ?

$8x+\frac{2}{3}$

$8\tfrac{2}{3}x^2$

Explanation

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{2-1})+(1*\frac{2}{3}x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{1})+(1*\frac{2}{3}x^{0})$

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(8x)+(1*\frac{2}{3}(1))$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=8x+\frac{2}{3}$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=?$

Explanation

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{2-1})+(0*8x^{0-1})$

Notice that $0*8x^{0-1}=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=12x$

What is the derivative of $2x^4+\frac{1}{2}x$ ?

$8x^{3}+\frac{1}{2}$

$8x^2+\frac{1}{2}x$

$\frac{2}{5}x^5+\frac{1}{4}x^2+c$

Explanation

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(4*2x^{4-1})+(1*\frac{1}{2}x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(4*2x^{3})+(\frac{1}{2}x^{0})$

Remember that anything to the zero power is equal to one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(8x^{3})+(\frac{1}{2}*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=8x^{3}+\frac{1}{2}$

Find the derivative of the following function at the point .

Explanation

Use the power rule on each term of the polynomial to get the derivative,

Now we plug in

What is the derivative of ?

$3x^2+4x+\frac{5}{x}$

$\frac{1}{4}x^4+\frac{2}{3}x^3+5x+c$

Explanation

To find the first derivative, we can use the power rule. To do that, we lower the exponent on the variables by one and multiply by the original exponent.

We're going to treat as since anything to the zero power is one.