Find the derivative of the function. f(x) = 7×2 ln(7x)
The Correct Answer and Explanation is:
To find the derivative of the function f(x)=7x2ln(7x)f(x) = 7x^2 \ln(7x)f(x)=7x2ln(7x), we’ll use the product rule since the function is the product of two functions: 7x27x^27×2 and ln(7x)\ln(7x)ln(7x). The product rule states: ddx[u(x)v(x)]=u′(x)v(x)+u(x)v′(x)\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)dxd[u(x)v(x)]=u′(x)v(x)+u(x)v′(x)
Here, u(x)=7x2u(x) = 7x^2u(x)=7×2 and v(x)=ln(7x)v(x) = \ln(7x)v(x)=ln(7x).
Step 1: Differentiate u(x)=7x2u(x) = 7x^2u(x)=7×2
The derivative of 7x27x^27×2 is straightforward. Using the power rule: u′(x)=14xu'(x) = 14xu′(x)=14x
Step 2: Differentiate v(x)=ln(7x)v(x) = \ln(7x)v(x)=ln(7x)
To differentiate ln(7x)\ln(7x)ln(7x), we need the chain rule. First, recognize that ln(7x)\ln(7x)ln(7x) can be written as ln(7)+ln(x)\ln(7) + \ln(x)ln(7)+ln(x). Since ln(7)\ln(7)ln(7) is a constant, its derivative is zero, and we only need to differentiate ln(x)\ln(x)ln(x). The derivative of ln(x)\ln(x)ln(x) is 1x\frac{1}{x}x1, so: v′(x)=17x×7=1xv'(x) = \frac{1}{7x} \times 7 = \frac{1}{x}v′(x)=7×1×7=x1
Step 3: Apply the Product Rule
Now, applying the product rule: f′(x)=u′(x)v(x)+u(x)v′(x)f'(x) = u'(x)v(x) + u(x)v'(x)f′(x)=u′(x)v(x)+u(x)v′(x)
Substitute the values of u(x)u(x)u(x), u′(x)u'(x)u′(x), v(x)v(x)v(x), and v′(x)v'(x)v′(x): f'(x) = (14x)(\ln(7x)) + (7x^2)\left(\frac{1}{x}\)
Simplifying the second term: f′(x)=14xln(7x)+7xf'(x) = 14x \ln(7x) + 7xf′(x)=14xln(7x)+7x
Final Answer
The derivative of the function is: f′(x)=14xln(7x)+7xf'(x) = 14x \ln(7x) + 7xf′(x)=14xln(7x)+7x
Explanation:
- Product Rule: Since the given function is a product of two functions 7x27x^27×2 and ln(7x)\ln(7x)ln(7x), we applied the product rule to differentiate it.
- Chain Rule: When differentiating ln(7x)\ln(7x)ln(7x), we used the chain rule to account for the inner function 7x7x7x.
- Simplification: After differentiating, we simplified the second term by canceling out xxx in the numerator and denominator.
This method allows us to break down the problem step by step and apply the necessary rules of differentiation.
