Example – Differentiation of a function w.r.t. Another Function

Practice Questions with Step-by-Step Solutions

Question 1

Find [\dfrac{dy}{du}] if [y=x^{3}+2x] and [u=x^{2}+1].

Step-by-step Solution:

1. Compute [\dfrac{dy}{dx}]: [\dfrac{d}{dx}\big(x^{3}+2x\big)][=3x^{2}+2].
2. Compute [\dfrac{du}{dx}]: [\dfrac{d}{dx}\big(x^{2}+1\big)][=2x].
3. Divide:
   [\dfrac{dy}{du}][=\dfrac{\dfrac{dy}{dx}}{\dfrac{du}{dx}}][=\dfrac{3x^{2}+2}{2x}], valid where [x\ne 0].

Conclusion:
[\dfrac{dy}{du}][=\dfrac{3x^{2}+2}{2x}].

Question 2

If [y=\sin x] and [u=\ln x], find [\dfrac{dy}{du}] for [x>0].

Step-by-step Solution:

1. [\dfrac{dy}{dx}=\cos x].
2. [\dfrac{du}{dx}=\dfrac{1}{x}].
3. Divide:
   [\dfrac{dy}{du}=\dfrac{\cos x}{1/x}][=x\cos x].

Conclusion:
[\dfrac{dy}{du}=x\cos x], valid for [x>0].

Question 3

Given [y=e^{2x}] and [u=\sin x], find [\dfrac{dy}{du}] where [\cos x\ne 0].

Step-by-step Solution:

1. [\dfrac{dy}{dx}=\dfrac{d}{dx}(e^{2x})][=2e^{2x}].
2. [\dfrac{du}{dx}=\cos x].
3. Divide:
   [\dfrac{dy}{du}=\dfrac{2e^{2x}}{\cos x}].

Conclusion:
[\dfrac{dy}{du}][=\dfrac{2e^{2x}}{\cos x}], provided [\cos x\ne 0].

Question 4

Find [\dfrac{dy}{du}] when [y=\ln(1+x)] and [u=x/(1+x)], [x>-1, x\ne -1].

Step-by-step Solution:

1. [\dfrac{dy}{dx}=\dfrac{1}{1+x}].
2. Compute [\dfrac{du}{dx}]:
   [u][=\dfrac{x}{1+x}][ \Rightarrow \dfrac{du}{dx}][=\dfrac{(1+x)\cdot 1 – x\cdot 1}{(1+x)^{2}}][=\dfrac{1}{(1+x)^{2}}].
3. Divide:
   [\dfrac{dy}{du}][=\dfrac{\dfrac{1}{1+x}}{\dfrac{1}{(1+x)^{2}}}][=\dfrac{1}{1+x}\cdot(1+x)^{2} = 1+x].

Conclusion:
[\dfrac{dy}{du}=1+x], domain [x\ne -1].

Question 5

If [y=\arctan x] and [u=x^{2}], find [\dfrac{dy}{du}] (stating domain).

Step-by-step Solution:

1. [\dfrac{dy}{dx}=\dfrac{1}{1+x^{2}}].
2. [\dfrac{du}{dx}=2x].
3. Divide:
   [\dfrac{dy}{du}][=\dfrac{1}{1+x^{2}}\cdot\dfrac{1}{2x}][=\dfrac{1}{2x(1+x^{2})}], valid where [x\ne 0].

Conclusion:
[\dfrac{dy}{du}=\dfrac{1}{2x(1+x^{2})}], [x\ne 0].

Question 6

Let [y=\sqrt{1+x^{2}}] and [u=\sin x]. Find [\dfrac{dy}{du}] where [\cos x\ne 0].

Step-by-step Solution:

1. [\dfrac{dy}{dx}][=\dfrac{1}{2}(1+x^{2})^{-1/2}\cdot 2x][ = \dfrac{x}{\sqrt{1+x^{2}}}].
2. [\dfrac{du}{dx}=\cos x].
3. Divide:
   [\dfrac{dy}{du}=\dfrac{x}{\sqrt{1+x^{2}};\cos x}].

Conclusion:
[\dfrac{dy}{du}][=\dfrac{x}{\cos x\sqrt{1+x^{2}}}], provided [\cos x\ne 0].

Question 7

Given [y=(x^{3}+1)^{1/2}] and [u=x^{2}+x], compute [\dfrac{dy}{du}] (specify domain where [du/dx\ne 0]).

Step-by-step Solution:

1. [\dfrac{dy}{dx}][=\dfrac{1}{2}(x^{3}+1)^{-1/2}\cdot 3x^{2}][ = \dfrac{3x^{2}}{2\sqrt{x^{3}+1}}].
2. [\dfrac{du}{dx}=2x+1].
3. Divide:
   [\dfrac{dy}{du}][=\dfrac{3x^{2}}{2\sqrt{x^{3}+1},(2x+1)}], valid where [2x+1\ne 0].

Conclusion:
[\dfrac{dy}{du}][=\dfrac{3x^{2}}{2(2x+1)\sqrt{x^{3}+1}}], for [x\ne -\tfrac{1}{2}].

Question 8

If [y=\sin(x^{2})] and [u=\cos x], find [\dfrac{dy}{du}] (state where defined).

Step-by-step Solution:

1. Use chain rule: [\dfrac{dy}{dx}][=\cos(x^{2})\cdot 2x][ = 2x\cos(x^{2})].
2. [\dfrac{du}{dx}=-\sin x].
3. Divide:
   [\dfrac{dy}{du}][=\dfrac{2x\cos(x^{2})}{-,\sin x}][ = -\dfrac{2x\cos(x^{2})}{\sin x}], valid where [\sin x\ne 0].

Conclusion:
[\dfrac{dy}{du}][=-\dfrac{2x\cos(x^{2})}{\sin x}], [\sin x\ne 0].

Question 9

Let [y=\ln(\sin x)] and [u=\tan x]. Find [\dfrac{dy}{du}] (state domain restrictions).

Step-by-step Solution:

1. [\dfrac{dy}{dx}=\dfrac{1}{\sin x}\cdot \cos x = \cot x], valid where [\sin x\ne 0].
2. [\dfrac{du}{dx}=\sec^{2}x], valid where [\cos x\ne 0].
3. Divide:
   [\dfrac{dy}{du}][=\dfrac{\cot x}{\sec^{2}x}][ = \cot x\cos^{2}x][ = \cos x\sin^{-1}x\cos^{2}x?]
   Better simplify: [\cot x = \dfrac{\cos x}{\sin x}], [\sec^{2}x=\dfrac{1}{\cos^{2}x}], so division gives:
   [\dfrac{\cos x/\sin x}{1/\cos^{2}x}][ = \dfrac{\cos^{3}x}{\sin x}].
4. Final result valid where [\sin x\ne 0] and [\cos x\ne 0].

Conclusion:
[\dfrac{dy}{du}][=\dfrac{\cos^{3}x}{\sin x}], for [\sin x\ne 0,\ \cos x\ne 0].

Question 10

If [y=\dfrac{1}{1+x^{2}}] and [u=\arctan x], compute [\dfrac{dy}{du}] (state domain).

Step-by-step Solution:

1. [\dfrac{dy}{dx}][=\dfrac{d}{dx}\big((1+x^{2})^{-1}\big)][ = -1\cdot(1+x^{2})^{-2}\cdot 2x][ = -\dfrac{2x}{(1+x^{2})^{2}}].
2. [\dfrac{du}{dx}=\dfrac{1}{1+x^{2}}].
3. Divide:
   [\dfrac{dy}{du}][=\dfrac{-\dfrac{2x}{(1+x^{2})^{2}}}{\dfrac{1}{1+x^{2}}}][ = -\dfrac{2x}{1+x^{2}}].
4. Domain: all real [x] (denominators nonzero for all real [x]).

Conclusion:
[\dfrac{dy}{du}][=-\dfrac{2x}{1+x^{2}}], valid for all real [x].