Let us solve the derivatives of the given functions step by step.
(i) ( 5 x + 3 ) − 3 (5x + 3)^{-3} ( 5 x + 3 ) − 3
To differentiate, we apply the chain rule :
d d x [ ( u ) n ] = n ⋅ u n − 1 ⋅ d u d x \frac{d}{dx}[(u)^n] = n \cdot u^{n-1} \cdot \frac{du}{dx} d x d [( u ) n ] = n ⋅ u n − 1 ⋅ d x d u
Here, u = 5 x + 3 u = 5x + 3 u = 5 x + 3 n = − 3 n = -3 n = − 3 d u d x = 5 \frac{du}{dx} = 5 d x d u = 5
Solution :
d d x [ ( 5 x + 3 ) − 3 ] = − 3 ⋅ ( 5 x + 3 ) − 4 ⋅ 5 = − 15 ⋅ ( 5 x + 3 ) − 4 \frac{d}{dx}[(5x + 3)^{-3}] = -3 \cdot (5x + 3)^{-4} \cdot 5 = -15 \cdot (5x + 3)^{-4} d x d [( 5 x + 3 ) − 3 ] = − 3 ⋅ ( 5 x + 3 ) − 4 ⋅ 5 = − 15 ⋅ ( 5 x + 3 ) − 4
Final Answer :
− 15 ⋅ ( 5 x + 3 ) − 4 \boxed{-15 \cdot (5x + 3)^{-4}} − 15 ⋅ ( 5 x + 3 ) − 4
(ii) ( 2 − 3 x ) 5 (2 - 3x)^5 ( 2 − 3 x ) 5
Here, we again use the chain rule :
d d x [ ( u ) n ] = n ⋅ u n − 1 ⋅ d u d x \frac{d}{dx}[(u)^n] = n \cdot u^{n-1} \cdot \frac{du}{dx} d x d [( u ) n ] = n ⋅ u n − 1 ⋅ d x d u
Here, u = 2 − 3 x u = 2 - 3x u = 2 − 3 x n = 5 n = 5 n = 5 d u d x = − 3 \frac{du}{dx} = -3 d x d u = − 3
Solution :
d d x [ ( 2 − 3 x ) 5 ] = 5 ⋅ ( 2 − 3 x ) 4 ⋅ ( − 3 ) = − 15 ⋅ ( 2 − 3 x ) 4 \frac{d}{dx}[(2 - 3x)^5] = 5 \cdot (2 - 3x)^4 \cdot (-3) = -15 \cdot (2 - 3x)^4 d x d [( 2 − 3 x ) 5 ] = 5 ⋅ ( 2 − 3 x ) 4 ⋅ ( − 3 ) = − 15 ⋅ ( 2 − 3 x ) 4
Final Answer :
− 15 ⋅ ( 2 − 3 x ) 4 \boxed{-15 \cdot (2 - 3x)^4} − 15 ⋅ ( 2 − 3 x ) 4
(iii) ( 3 + 7 x ) − 5 / 3 (3 + 7x)^{-5/3} ( 3 + 7 x ) − 5/3
Here, we use the chain rule :
d d x [ ( u ) n ] = n ⋅ u n − 1 ⋅ d u d x \frac{d}{dx}[(u)^n] = n \cdot u^{n-1} \cdot \frac{du}{dx} d x d [( u ) n ] = n ⋅ u n − 1 ⋅ d x d u
Here, u = 3 + 7 x u = 3 + 7x u = 3 + 7 x n = − 5 3 n = -\frac{5}{3} n = − 3 5 d u d x = 7 \frac{du}{dx} = 7 d x d u = 7
Solution :
d d x [ ( 3 + 7 x ) − 5 / 3 ] = − 5 3 ⋅ ( 3 + 7 x ) − 8 / 3 ⋅ 7 = − 35 3 ⋅ ( 3 + 7 x ) − 8 / 3 \frac{d}{dx}[(3 + 7x)^{-5/3}] = -\frac{5}{3} \cdot (3 + 7x)^{-8/3} \cdot 7 = -\frac{35}{3} \cdot (3 + 7x)^{-8/3} d x d [( 3 + 7 x ) − 5/3 ] = − 3 5 ⋅ ( 3 + 7 x ) − 8/3 ⋅ 7 = − 3 35 ⋅ ( 3 + 7 x ) − 8/3
Final Answer :
− 35 3 ⋅ ( 3 + 7 x ) − 8 / 3 \boxed{-\frac{35}{3} \cdot (3 + 7x)^{-8/3}} − 3 35 ⋅ ( 3 + 7 x ) − 8/3
(iv) 2 x + 5 \sqrt{2x + 5} 2 x + 5
This is equivalent to ( 2 x + 5 ) 1 / 2 (2x + 5)^{1/2} ( 2 x + 5 ) 1/2 chain rule again:
d d x [ ( u ) n ] = n ⋅ u n − 1 ⋅ d u d x \frac{d}{dx}[(u)^n] = n \cdot u^{n-1} \cdot \frac{du}{dx} d x d [( u ) n ] = n ⋅ u n − 1 ⋅ d x d u
Here, u = 2 x + 5 u = 2x + 5 u = 2 x + 5 n = 1 2 n = \frac{1}{2} n = 2 1 d u d x = 2 \frac{du}{dx} = 2 d x d u = 2
Solution :
d d x [ 2 x + 5 ] = 1 2 ⋅ ( 2 x + 5 ) − 1 / 2 ⋅ 2 = ( 2 x + 5 ) − 1 / 2 \frac{d}{dx}[\sqrt{2x + 5}] = \frac{1}{2} \cdot (2x + 5)^{-1/2} \cdot 2 = (2x + 5)^{-1/2} d x d [ 2 x + 5 ] = 2 1 ⋅ ( 2 x + 5 ) − 1/2 ⋅ 2 = ( 2 x + 5 ) − 1/2
Final Answer :
1 2 x + 5 \boxed{\frac{1}{\sqrt{2x + 5}}} 2 x + 5 1
(v) 1 2 − 9 x \frac{1}{\sqrt{2 - 9x}} 2 − 9 x 1
This is equivalent to ( 2 − 9 x ) − 1 / 2 (2 - 9x)^{-1/2} ( 2 − 9 x ) − 1/2 chain rule :
d d x [ ( u ) n ] = n ⋅ u n − 1 ⋅ d u d x \frac{d}{dx}[(u)^n] = n \cdot u^{n-1} \cdot \frac{du}{dx} d x d [( u ) n ] = n ⋅ u n − 1 ⋅ d x d u
Here, u = 2 − 9 x u = 2 - 9x u = 2 − 9 x n = − 1 2 n = -\frac{1}{2} n = − 2 1 d u d x = − 9 \frac{du}{dx} = -9 d x d u = − 9
Solution :
d d x [ ( 2 − 9 x ) − 1 / 2 ] = − 1 2 ⋅ ( 2 − 9 x ) − 3 / 2 ⋅ ( − 9 ) = 9 2 ⋅ ( 2 − 9 x ) − 3 / 2 \frac{d}{dx}[(2 - 9x)^{-1/2}] = -\frac{1}{2} \cdot (2 - 9x)^{-3/2} \cdot (-9) = \frac{9}{2} \cdot (2 - 9x)^{-3/2} d x d [( 2 − 9 x ) − 1/2 ] = − 2 1 ⋅ ( 2 − 9 x ) − 3/2 ⋅ ( − 9 ) = 2 9 ⋅ ( 2 − 9 x ) − 3/2
Final Answer :
9 2 ⋅ ( 2 − 9 x ) − 3 / 2 \boxed{\frac{9}{2} \cdot (2 - 9x)^{-3/2}} 2 9 ⋅ ( 2 − 9 x ) − 3/2
Let me know if you need further clarification or detailed steps. 😊
Follow-Up Questions:
How is the chain rule applied in these problems?
Why does the power decrease by 1 in derivative calculations?
How can you rewrite negative exponents in simpler forms?
What happens when you differentiate a square root function?
How do these rules apply to composite functions in general?
Tip:
Always simplify the function into a power/exponent form before applying the chain rule for easier computation.