test iii. fundamentals\n5. find $\frac{dy}{dx}$ when $axy=\frac{sqrt{3x - 2}}{sqrt{4x^{2}+3}}$ where a is…

test iii. fundamentals\n5. find $\frac{dy}{dx}$ when $axy=\frac{sqrt{3x - 2}}{sqrt{4x^{2}+3}}$ where a is constant.\n6. find $\frac{dd}{dc}$ when $abcd=\frac{-b + sqrt{b^{2}-4ac}}{2a}$ where a,b,c,d are not constant.\n7. find $\frac{dy}{dx}$ when $y^{4}=\frac{1}{sqrt6{x^{3}}}cdot\frac{x}{x + 2}$\n8. find $\frac{dz}{dx}$ when $sqrt{y}=3x^{5}-sqrt{4(x + 2)^{2}-2}$

test iii. fundamentals\n5. find $\frac{dy}{dx}$ when $axy=\frac{sqrt{3x - 2}}{sqrt{4x^{2}+3}}$ where a is constant.\n6. find $\frac{dd}{dc}$ when $abcd=\frac{-b + sqrt{b^{2}-4ac}}{2a}$ where a,b,c,d are not constant.\n7. find $\frac{dy}{dx}$ when $y^{4}=\frac{1}{sqrt6{x^{3}}}cdot\frac{x}{x + 2}$\n8. find $\frac{dz}{dx}$ when $sqrt{y}=3x^{5}-sqrt{4(x + 2)^{2}-2}$

Answer

Explanation:

Step1: Differentiate implicitly for problem 5

Given (axy=\frac{\sqrt{3x - 2}}{\sqrt{4x^{2}+3}}), use the product - rule ((uv)^\prime = u^\prime v+uv^\prime) on the left - hand side ((u = ax), (v = y)) and the quotient - rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}) on the right - hand side ((u=\sqrt{3x - 2}=(3x - 2)^{\frac{1}{2}}), (v=\sqrt{4x^{2}+3}=(4x^{2}+3)^{\frac{1}{2}})). Left - hand side: ((axy)^\prime=a(y + x\frac{dy}{dx})). Right - hand side: [ \begin{align*} u^\prime&=\frac{3}{2\sqrt{3x - 2}}\ v^\prime&=\frac{8x}{2\sqrt{4x^{2}+3}}=\frac{4x}{\sqrt{4x^{2}+3}}\ \left(\frac{\sqrt{3x - 2}}{\sqrt{4x^{2}+3}}\right)^\prime&=\frac{\frac{3}{2\sqrt{3x - 2}}\sqrt{4x^{2}+3}-\sqrt{3x - 2}\frac{4x}{\sqrt{4x^{2}+3}}}{4x^{2}+3}\ &=\frac{3(4x^{2}+3)-8x(3x - 2)}{2(4x^{2}+3)^{\frac{3}{2}}\sqrt{3x - 2}}\ &=\frac{12x^{2}+9-24x^{2}+16x}{2(4x^{2}+3)^{\frac{3}{2}}\sqrt{3x - 2}}\ &=\frac{- 12x^{2}+16x + 9}{2(4x^{2}+3)^{\frac{3}{2}}\sqrt{3x - 2}} \end{align*} ] Then (a(y + x\frac{dy}{dx})=\frac{- 12x^{2}+16x + 9}{2(4x^{2}+3)^{\frac{3}{2}}\sqrt{3x - 2}}), and (\frac{dy}{dx}=\frac{\frac{- 12x^{2}+16x + 9}{2a(4x^{2}+3)^{\frac{3}{2}}\sqrt{3x - 2}}-y}{x}).

Step2: Differentiate implicitly for problem 6

The equation (abcd=\frac{-b+\sqrt{b^{2}-4ac}}{2a}) is not a function of (x) in a way that we can find (\frac{dd}{dc}) as presented. It seems there is a mis - statement. If we assume it is a wrong equation and we should consider a different context, let's assume it is a non - related error and move on.

Step3: Differentiate implicitly for problem 7

Given (y^{4}=\frac{1}{\sqrt{x}}+\frac{x}{2}), rewrite it as (y^{4}=x^{-\frac{1}{2}}+\frac{x}{2}). Differentiate both sides with respect to (x): (4y^{3}\frac{dy}{dx}=-\frac{1}{2}x^{-\frac{3}{2}}+\frac{1}{2}). Then (\frac{dy}{dx}=\frac{-\frac{1}{2}x^{-\frac{3}{2}}+\frac{1}{2}}{4y^{3}}=\frac{-1 + x^{\frac{3}{2}}}{8y^{3}x^{\frac{3}{2}}}).

Step4: Differentiate implicitly for problem 8

Given (\sqrt{y}=3x^{5}-\sqrt{4(x + 2)^{2}-2}), rewrite it as (y^{\frac{1}{2}}=3x^{5}-\sqrt{4(x^{2}+4x + 4)-2}=3x^{5}-\sqrt{4x^{2}+16x + 14}). Differentiate both sides with respect to (x): (\frac{1}{2\sqrt{y}}\frac{dy}{dx}=15x^{4}-\frac{8x + 16}{2\sqrt{4x^{2}+16x + 14}}). Then (\frac{dy}{dx}=2\sqrt{y}\left(15x^{4}-\frac{4x + 8}{\sqrt{4x^{2}+16x + 14}}\right)).

Answer:

For problem 5: (\frac{dy}{dx}=\frac{\frac{- 12x^{2}+16x + 9}{2a(4x^{2}+3)^{\frac{3}{2}}\sqrt{3x - 2}}-y}{x}) For problem 7: (\frac{dy}{dx}=\frac{-1 + x^{\frac{3}{2}}}{8y^{3}x^{\frac{3}{2}}}) For problem 8: (\frac{dy}{dx}=2\sqrt{y}\left(15x^{4}-\frac{4x + 8}{\sqrt{4x^{2}+16x + 14}}\right)) (Problem 6 seems to have an incorrect equation for the given operation)