What is the derivative of the function y = tan (x^2 + 3x). A. y’ = (2x + 3) sec^2 (x^2 + 3x) B. y’ = (2x – 3) sec^2 (x^2 + 3x) C. y’ = (2x + 3) sec^2 (x^2 – 3x) D. y’ = (2x – 3) sec^2 (x^2 – 3x)
The Correct Answer and Explanation is:
The function you are working with is: y=tan(x2+3x)y = \tan(x^2 + 3x)y=tan(x2+3x)
To find the derivative of this function, we need to apply the chain rule, which is used when differentiating compositions of functions. Here, we have two functions:
- The outer function: tan(u)\tan(u)tan(u), where u=x2+3xu = x^2 + 3xu=x2+3x
- The inner function: u=x2+3xu = x^2 + 3xu=x2+3x
Step 1: Differentiate the outer function with respect to uuu
The derivative of tan(u)\tan(u)tan(u) is: ddu(tan(u))=sec2(u)\frac{d}{du} \left( \tan(u) \right) = \sec^2(u)dud(tan(u))=sec2(u)
Step 2: Differentiate the inner function u=x2+3xu = x^2 + 3xu=x2+3x with respect to xxx
The derivative of x2+3xx^2 + 3xx2+3x is: ddx(x2+3x)=2x+3\frac{d}{dx} \left( x^2 + 3x \right) = 2x + 3dxd(x2+3x)=2x+3
Step 3: Apply the chain rule
The derivative of y=tan(x2+3x)y = \tan(x^2 + 3x)y=tan(x2+3x) with respect to xxx is: dydx=ddu(tan(u))⋅dudx\frac{dy}{dx} = \frac{d}{du} \left( \tan(u) \right) \cdot \frac{du}{dx}dxdy=dud(tan(u))⋅dxdu
Substituting the derivatives from Steps 1 and 2: y′=sec2(x2+3x)⋅(2x+3)y’ = \sec^2(x^2 + 3x) \cdot (2x + 3)y′=sec2(x2+3x)⋅(2x+3)
Thus, the correct derivative is: y′=(2x+3)sec2(x2+3x)y’ = (2x + 3) \sec^2(x^2 + 3x)y′=(2x+3)sec2(x2+3x)
Conclusion:
The correct answer is A. y′=(2x+3)sec2(x2+3x)y’ = (2x + 3) \sec^2(x^2 + 3x)y′=(2x+3)sec2(x2+3x).
Explanation:
- The chain rule is essential here because the function involves the composition of tan(u)\tan(u)tan(u) and u=x2+3xu = x^2 + 3xu=x2+3x.
- We first differentiate the outer function tan(u)\tan(u)tan(u), yielding sec2(u)\sec^2(u)sec2(u), and then differentiate the inner function x2+3xx^2 + 3xx2+3x, which gives 2x+32x + 32x+3.
- Combining these two results gives the derivative as y′=(2x+3)sec2(x2+3x)y’ = (2x + 3) \sec^2(x^2 + 3x)y′=(2x+3)sec2(x2+3x).
