Find dx/dy y=(4x+3)^5

$y = {(4 x + 3)}^{5}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (y) = \frac{d}{d y} ({(4 x + 3)}^{5})$

Step 2

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$1$

Step 3

Differentiate the right side of the equation.

Use the Binomial Theorem.

$\frac{d}{d y} [{(4 x)}^{5} + 5 {(4 x)}^{4} \cdot 3 + 10 {(4 x)}^{3} \cdot 3^{2} + 10 {(4 x)}^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Differentiate.

Simplify each term.

Apply the product rule to $4 x$ .

$\frac{d}{d y} [4^{5} x^{5} + 5 {(4 x)}^{4} \cdot 3 + 10 {(4 x)}^{3} \cdot 3^{2} + 10 {(4 x)}^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Raise $4$ to the power of $5$ .

$\frac{d}{d y} [1024 x^{5} + 5 {(4 x)}^{4} \cdot 3 + 10 {(4 x)}^{3} \cdot 3^{2} + 10 {(4 x)}^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Apply the product rule to $4 x$ .

$\frac{d}{d y} [1024 x^{5} + 5 (4^{4} x^{4}) \cdot 3 + 10 {(4 x)}^{3} \cdot 3^{2} + 10 {(4 x)}^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Raise $4$ to the power of $4$ .

$\frac{d}{d y} [1024 x^{5} + 5 (256 x^{4}) \cdot 3 + 10 {(4 x)}^{3} \cdot 3^{2} + 10 {(4 x)}^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Multiply $256$ by $5$ .

$\frac{d}{d y} [1024 x^{5} + 1280 x^{4} \cdot 3 + 10 {(4 x)}^{3} \cdot 3^{2} + 10 {(4 x)}^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Multiply $3$ by $1280$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 10 {(4 x)}^{3} \cdot 3^{2} + 10 {(4 x)}^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Apply the product rule to $4 x$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 10 (4^{3} x^{3}) \cdot 3^{2} + 10 {(4 x)}^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Raise $4$ to the power of $3$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 10 (64 x^{3}) \cdot 3^{2} + 10 {(4 x)}^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Multiply $64$ by $10$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 640 x^{3} \cdot 3^{2} + 10 {(4 x)}^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Raise $3$ to the power of $2$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 640 x^{3} \cdot 9 + 10 {(4 x)}^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Multiply $9$ by $640$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 5760 x^{3} + 10 {(4 x)}^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Apply the product rule to $4 x$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 5760 x^{3} + 10 (4^{2} x^{2}) \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Raise $4$ to the power of $2$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 5760 x^{3} + 10 (16 x^{2}) \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Multiply $16$ by $10$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 5760 x^{3} + 160 x^{2} \cdot 3^{3} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Raise $3$ to the power of $3$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 5760 x^{3} + 160 x^{2} \cdot 27 + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Multiply $27$ by $160$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 5760 x^{3} + 4320 x^{2} + 5 (4 x) \cdot 3^{4} + 3^{5}]$

Multiply $4$ by $5$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 5760 x^{3} + 4320 x^{2} + 20 x \cdot 3^{4} + 3^{5}]$

Raise $3$ to the power of $4$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 5760 x^{3} + 4320 x^{2} + 20 x \cdot 81 + 3^{5}]$

Multiply $81$ by $20$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 5760 x^{3} + 4320 x^{2} + 1620 x + 3^{5}]$

Raise $3$ to the power of $5$ .

$\frac{d}{d y} [1024 x^{5} + 3840 x^{4} + 5760 x^{3} + 4320 x^{2} + 1620 x + 243]$

By the Sum Rule, the derivative of $1024 x^{5} + 3840 x^{4} + 5760 x^{3} + 4320 x^{2} + 1620 x + 243$ with respect to $y$ is $\frac{d}{d y} [1024 x^{5}] + \frac{d}{d y} [3840 x^{4}] + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$ .

$\frac{d}{d y} [1024 x^{5}] + \frac{d}{d y} [3840 x^{4}] + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Since $1024$ is constant with respect to $y$ , the derivative of $1024 x^{5}$ with respect to $y$ is $1024 \frac{d}{d y} [x^{5}]$ .

$1024 \frac{d}{d y} [x^{5}] + \frac{d}{d y} [3840 x^{4}] + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{5}$ and $g (y) = x$ .

To apply the Chain Rule, set $u_{1}$ as $x$ .

$1024 (\frac{d}{d u_{1}} [{u_{1}}^{5}] \frac{d}{d y} [x]) + \frac{d}{d y} [3840 x^{4}] + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 5$ .

$1024 (5 {u_{1}}^{4} \frac{d}{d y} [x]) + \frac{d}{d y} [3840 x^{4}] + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Replace all occurrences of $u_{1}$ with $x$ .

$1024 (5 x^{4} \frac{d}{d y} [x]) + \frac{d}{d y} [3840 x^{4}] + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Multiply $5$ by $1024$ .

$5120 (x^{4} \frac{d}{d y} [x]) + \frac{d}{d y} [3840 x^{4}] + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Rewrite $\frac{d}{d y} [x]$ as $x'$ .

$5120 x^{4} x' + \frac{d}{d y} [3840 x^{4}] + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Since $3840$ is constant with respect to $y$ , the derivative of $3840 x^{4}$ with respect to $y$ is $3840 \frac{d}{d y} [x^{4}]$ .

$5120 x^{4} x' + 3840 \frac{d}{d y} [x^{4}] + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{4}$ and $g (y) = x$ .

To apply the Chain Rule, set $u_{2}$ as $x$ .

$5120 x^{4} x' + 3840 (\frac{d}{d u_{2}} [{u_{2}}^{4}] \frac{d}{d y} [x]) + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 4$ .

$5120 x^{4} x' + 3840 (4 {u_{2}}^{3} \frac{d}{d y} [x]) + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Replace all occurrences of $u_{2}$ with $x$ .

$5120 x^{4} x' + 3840 (4 x^{3} \frac{d}{d y} [x]) + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Multiply $4$ by $3840$ .

$5120 x^{4} x' + 15360 (x^{3} \frac{d}{d y} [x]) + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Rewrite $\frac{d}{d y} [x]$ as $x'$ .

$5120 x^{4} x' + 15360 x^{3} x' + \frac{d}{d y} [5760 x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Since $5760$ is constant with respect to $y$ , the derivative of $5760 x^{3}$ with respect to $y$ is $5760 \frac{d}{d y} [x^{3}]$ .

$5120 x^{4} x' + 15360 x^{3} x' + 5760 \frac{d}{d y} [x^{3}] + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{3}$ and $g (y) = x$ .

To apply the Chain Rule, set $u_{3}$ as $x$ .

$5120 x^{4} x' + 15360 x^{3} x' + 5760 (\frac{d}{d u_{3}} [{u_{3}}^{3}] \frac{d}{d y} [x]) + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Differentiate using the Power Rule which states that $\frac{d}{d u_{3}} [{u_{3}}^{n}]$ is $n {u_{3}}^{n - 1}$ where $n = 3$ .

$5120 x^{4} x' + 15360 x^{3} x' + 5760 (3 {u_{3}}^{2} \frac{d}{d y} [x]) + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Replace all occurrences of $u_{3}$ with $x$ .

$5120 x^{4} x' + 15360 x^{3} x' + 5760 (3 x^{2} \frac{d}{d y} [x]) + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Multiply $3$ by $5760$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 (x^{2} \frac{d}{d y} [x]) + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Rewrite $\frac{d}{d y} [x]$ as $x'$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + \frac{d}{d y} [4320 x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Since $4320$ is constant with respect to $y$ , the derivative of $4320 x^{2}$ with respect to $y$ is $4320 \frac{d}{d y} [x^{2}]$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 4320 \frac{d}{d y} [x^{2}] + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{2}$ and $g (y) = x$ .

To apply the Chain Rule, set $u_{4}$ as $x$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 4320 (\frac{d}{d u_{4}} [{u_{4}}^{2}] \frac{d}{d y} [x]) + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Differentiate using the Power Rule which states that $\frac{d}{d u_{4}} [{u_{4}}^{n}]$ is $n {u_{4}}^{n - 1}$ where $n = 2$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 4320 (2 u_{4} \frac{d}{d y} [x]) + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Replace all occurrences of $u_{4}$ with $x$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 4320 (2 x \frac{d}{d y} [x]) + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Multiply $2$ by $4320$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 8640 (x \frac{d}{d y} [x]) + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Rewrite $\frac{d}{d y} [x]$ as $x'$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 8640 x x' + \frac{d}{d y} [1620 x] + \frac{d}{d y} [243]$

Since $1620$ is constant with respect to $y$ , the derivative of $1620 x$ with respect to $y$ is $1620 \frac{d}{d y} [x]$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 8640 x x' + 1620 \frac{d}{d y} [x] + \frac{d}{d y} [243]$

Rewrite $\frac{d}{d y} [x]$ as $x'$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 8640 x x' + 1620 x' + \frac{d}{d y} [243]$

Since $243$ is constant with respect to $y$ , the derivative of $243$ with respect to $y$ is $0$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 8640 x x' + 1620 x' + 0$

Add $5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 8640 x x' + 1620 x'$ and $0$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 8640 x x' + 1620 x'$

Step 4

Reform the equation by setting the left side equal to the right side.

$1 = 5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 8640 x x' + 1620 x'$

Step 5

Solve for $x'$ .

Rewrite the equation as $5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 8640 x x' + 1620 x' = 1$ .

$5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 8640 x x' + 1620 x' = 1$

Factor $20 x'$ out of $5120 x^{4} x' + 15360 x^{3} x' + 17280 x^{2} x' + 8640 x x' + 1620 x'$ .

Factor $20 x'$ out of $5120 x^{4} x'$ .

$20 x' (256 x^{4}) + 15360 x^{3} x' + 17280 x^{2} x' + 8640 x x' + 1620 x' = 1$

Factor $20 x'$ out of $15360 x^{3} x'$ .

$20 x' (256 x^{4}) + 20 x' (768 x^{3}) + 17280 x^{2} x' + 8640 x x' + 1620 x' = 1$

Factor $20 x'$ out of $17280 x^{2} x'$ .

$20 x' (256 x^{4}) + 20 x' (768 x^{3}) + 20 x' (864 x^{2}) + 8640 x x' + 1620 x' = 1$

Factor $20 x'$ out of $8640 x x'$ .

$20 x' (256 x^{4}) + 20 x' (768 x^{3}) + 20 x' (864 x^{2}) + 20 x' (432 x) + 1620 x' = 1$

Factor $20 x'$ out of $1620 x'$ .

$20 x' (256 x^{4}) + 20 x' (768 x^{3}) + 20 x' (864 x^{2}) + 20 x' (432 x) + 20 x' \cdot 81 = 1$

Factor $20 x'$ out of $20 x' (256 x^{4}) + 20 x' (768 x^{3})$ .

$20 x' (256 x^{4} + 768 x^{3}) + 20 x' (864 x^{2}) + 20 x' (432 x) + 20 x' \cdot 81 = 1$

Factor $20 x'$ out of $20 x' (256 x^{4} + 768 x^{3}) + 20 x' (864 x^{2})$ .

$20 x' (256 x^{4} + 768 x^{3} + 864 x^{2}) + 20 x' (432 x) + 20 x' \cdot 81 = 1$

Factor $20 x'$ out of $20 x' (256 x^{4} + 768 x^{3} + 864 x^{2}) + 20 x' (432 x)$ .

$20 x' (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x) + 20 x' \cdot 81 = 1$

Factor $20 x'$ out of $20 x' (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x) + 20 x' \cdot 81$ .

$20 x' (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81) = 1$

Divide each term in $20 x' (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81) = 1$ by $20 (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)$ and simplify.

Divide each term in $20 x' (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81) = 1$ by $20 (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)$ .

$\frac{20 x' (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)}{20 (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)} = \frac{1}{20 (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)}$

Simplify the left side.

Cancel the common factor of $20$ .

Cancel the common factor.

$\frac{20 x' (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)}{20 (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)} = \frac{1}{20 (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)}$

Rewrite the expression.

$\frac{x' (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)}{256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81} = \frac{1}{20 (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)}$

Cancel the common factor of $256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81$ .

Cancel the common factor.

$\frac{x' (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)}{256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81} = \frac{1}{20 (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)}$

Divide $x'$ by $1$ .

$x' = \frac{1}{20 (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)}$

Step 6

Replace $x'$ with $\frac{d x}{d y}$ .

$\frac{d x}{d y} = \frac{1}{20 (256 x^{4} + 768 x^{3} + 864 x^{2} + 432 x + 81)}$

Find dx/dy y=(4x+3)^5

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