What is the derivative of ${tan (4 x)}^{tan (5 x)}$ $tan(4x)^tan(5x)$ ?

Question

What is the derivative of ${tan (4 x)}^{tan (5 x)}$ $tan(4x)^tan(5x)$ ?

1 Answer

Impact of this question

2104 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-24T13:35:48

We have the function

$y = {tan (4 x)}^{tan (5 x)}$ $y=tan(4x)^tan(5x)$

Take the natural logarithm of both sides:

$ln (y) = ln ({tan (4 x)}^{tan (5 x)})$ $ln(y)=ln(tan(4x)^tan(5x))$

Using the rule $ln (a^{b}) = b ln (a)$ $ln(a^b)=bln(a)$ , rewrite the right hand side:

$ln (y) = tan (5 x) \cdot ln (tan (4 x))$ $ln(y)=tan(5x)*ln(tan(4x))$

Differentiate both sides. The chain rule will be in effect on the left hand side, and primarily we will use the product rule on the right hand side.

$\frac{d y}{d x} (\frac{1}{y}) = ln (tan (4 x)) \frac{d}{d x} tan (5 x) + tan (5 x) \frac{d}{d x} ln (tan (4 x))$ $dy/dx(1/y)=ln(tan(4x))d/dxtan(5x)+tan(5x)d/dxln(tan(4x))$

Differentiate each, again using the chain rule.

$\frac{d y}{d x} (\frac{1}{y}) = 5 {sec}^{2} (5 x) ln (tan (4 x)) + tan (5 x) (\frac{4 {sec}^{2} (4 x)}{tan (4 x)})$ $dy/dx(1/y)=5sec^2(5x)ln(tan(4x))+tan(5x)((4sec^2(4x))/tan(4x))$

Multiply this all by $y$ $y$ , which equals ${tan (4 x)}^{tan (5 x)}$ $tan(4x)^tan(5x)$ , to solve for $\frac{d y}{d x}$ $dy/dx$ .

$\frac{d y}{d x} = {tan (4 x)}^{tan (5 x)} (5 {sec}^{2} (5 x) ln (tan (4 x)) + \frac{4 {sec}^{2} (4 x) tan (5 x)}{tan (4 x)})$ $dy/dx=tan(4x)^tan(5x)(5sec^2(5x)ln(tan(4x))+(4sec^2(4x)tan(5x))/tan(4x))$

Science

Math

Humanities

... and beyond

What is the derivative of ${tan (4 x)}^{tan (5 x)}$ $tan(4x)^tan(5x)$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the derivative of tan(4x)tan(5x) tan(4x)^tan(5x)?

1 Answer

Related questions

Impact of this question

What is the derivative of ${tan (4 x)}^{tan (5 x)}$ $tan(4x)^tan(5x)$ ?