Second Derivative of a Function – Examples with answers

To find the second derivative, we have to start by finding the first derivative of the function. Thus, we differentiate the function using the power rule of derivatives:

$latex f(x)=5x^3$

$latex f'(x)=15x^2$

Now, we use the power rule again to differentiate $latex f'(x)$:

$latex f'(x)=15x^2$

$latex f^{\prime \prime}(x)=30x$

The second derivative is $latex f^{\prime \prime}(x)=30x$.

We can use the power rule of derivatives on each term of the function to find the first derivative:

$latex f(x)=4x^4-2x^2+7x$

$latex f'(x)=16x^3-4x+7$

Now, let’s differentiate $latex f'(x)$ to find the second derivative using the same method:

$latex f'(x)=16x^3-4x+7$

$latex f^{\prime \prime}(x)=48x^2-4$

We have a rational expression. Therefore, we can use the laws of exponents to write as follows:

$$f(x)=5x^6+\frac{3}{x}$$

$latex f(x)=5x^6+3x^{-1}$

Now that we only have numerical exponents, we use the power rule to differentiate:

$latex f(x)=5x^6+3x^{-1}$

$latex f'(x)=30x^5-3x^{-2}$

Differentiating $latex f'(x)$, we have:

$latex f'(x)=30x^5-3x^{-2}$

$latex f^{\prime \prime}(x)=150x^4+6x^{-3}$

Using the laws of exponents, we can write as follows:

$latex f^{\prime \prime}(x)=150x^4+6x^{-3}$

$$f^{\prime \prime}(x)=150x^4+\frac{6}{x^3}$$

We can write the square root as a numerical exponent:

$$f(x) = -3x^{-5}+\sqrt{x}$$

$$f(x)=-3x^{-5}+x^{\frac{1}{2}}$$

Now, we can find the first derivative as follows:

$$f(x)=-3x^{-5}+x^{\frac{1}{2}}$$

$$f'(x)=15x^{-6}+\frac{1}{2}x^{-\frac{1}{2}}$$

We find the second derivative by differentiating $latex f'(x)$:

$$f'(x)=15x^{-6}+\frac{1}{2}x^{-\frac{1}{2}}$$

$$f^{\prime \prime}(x)=-90x^{-7}-\frac{1}{4}x^{-\frac{3}{2}}$$

We can simplify by using the laws of exponents and writing as follows:

$$f^{\prime \prime}(x)=-90x^{-7}-\frac{1}{4}x^{-\frac{3}{2}}$$

$$f^{\prime \prime}(x)=-\frac{90}{x^7}-\frac{1}{4\sqrt{x^3}}$$

In this case, we have a trigonometric function. We can differentiate by recalling that the derivative of sine equals cosine and the derivative of cosine equals negative sine:

$latex f(x)=\sin(x)-\cos(x)$

$latex f'(x)=\cos(x)+\sin(x)$

We apply the same rules to derive again:

$latex f'(x)=\cos(x)+\sin(x)$

$latex f^{\prime \prime}(x)=-\sin(x)+\cos(x)$

We can start by simplifying as follows:

$$f(x)=\frac{1}{\sqrt{x}}+3x^{-3}+5$$

$latex f(x)=x^{-\frac{1}{2}}+3x^{-3}+5$

Now, we differentiate using the power rule on each term of the function:

$latex f(x)=x^{-\frac{1}{2}}+3x^{-3}+5$

$$f'(x)=-\frac{1}{2}x^{-\frac{3}{2}}-9x^{-4}$$

We differentiate $latex f'(x)$ to obtain the second derivative:

$$f'(x)=-\frac{1}{2}x^{-\frac{3}{2}}-9x^{-4}$$

$$ f”(x)=\frac{3}{4}x^{-\frac{5}{2}}+36x^{-5}$$

We use the laws of exponents again to simplify:

$$ f^{\prime \prime}(x)=\frac{3}{4}x^{-\frac{5}{2}}+36x^{-5}$$

$$ f^{\prime \prime}(x)=\frac{3}{4x^{\frac{5}{2}}}+\frac{36}{x^5}$$

$$ f^{\prime \prime}(x)=\frac{3}{4\sqrt{x^5}}+\frac{36}{x^5}$$

We use the laws of exponents to write like this:

$$f(x)=\frac{2}{3x^2}-\frac{4}{x^5}$$

$$ f(x)=\frac{2}{3}x^{-2}-4x^{-5}$$

Using the power rule on both terms of the function, we have:

$$f(x)=\frac{2}{3}x^{-2}-4x^{-5}$$

$$f'(x)=-\frac{4}{3}x^{-3}+20x^{-6}$$

Differentiating $latex f'(x)$, we have:

$$f'(x)=-\frac{4}{3}x^{-3}+20x^{-6}$$

$$f^{\prime \prime}(x)=4x^{-4}-120x^{-7}$$

Finally, we can write as follows:

$$f^{\prime \prime}(x)= \frac{4}{x^4}-\frac{120}{x^7}$$

To solve this problem, we have to start by finding the first and second derivatives of the function. Thus, we have:

$latex f(x)=4x^3$

$latex f'(x)=12x^2$

$latex f^{\prime \prime}(x)=24x$

Now, we can substitute into the left-hand side of $latex 3f(x)f^{\prime \prime}(x)-2f'(x)^2=0$, and we have:

$latex 3f(x)f^{\prime \prime}(x)-2f'(x)^2=0$

$$3(4x^3)(24x)-2(12x^2)^2=0$$

$latex 288x^4-288x^4=0$

We see that the left side is equal to 0, so we have proved what is required.

Similar to the previous example, we need to start by finding the first and second derivatives of the function. Thus, we have:

$$f(x)=\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$$

$$f'(x)=-\frac{1}{2}x^{-\frac{3}{2}}$$

$$f^{\prime \prime}(x)=\frac{3}{4}x^{-\frac{5}{2}}$$

Now that we have the derivatives, let’s substitute in the left-hand side of $latex 2xf^{\prime \prime}(x)+3f'(x)=0$, and we have:

$latex 2xf^{\prime \prime}(x)+3f'(x)=0$

$$2x(\frac{3}{4}x^{-\frac{5}{2}})+3(-\frac{1}{2}x^{-\frac{3}{2}})=0$$

$$\frac{3}{2}x^{-\frac{3}{2}}-\frac{3}{2}x^{-\frac{3}{2}}=0$$

The left-hand side is equal to 0, so we have proved that the given expression is true.

The derivatives of the given function are as follows:

$latex f(x)=x^4$

$latex f'(x)=4x^3$

$latex f^{\prime \prime}(x)=12x^2$

Now, we can substitute into the left-hand side of $latex \frac{4y}{3}f^{\prime \prime}(x)-(f'(x))^2=0$, and we have:

$$\frac{4y}{3}f^{\prime \prime}(x)-(f'(x))^2=0$$

$$\frac{4(x^4)}{3}(12x^2)-(4x^3)^2=0$$

$latex 16x^6-16x^6=0$

We see that the left-hand side is equal to 0, so we have proved that the given expression is true.