# Let x(t) = t - t^3 and y(t) = t^3. Suppose that \frac{\partial g}{\partial x} = 5 and...

## Question:

Let {eq}\displaystyle x(t) = t - t^3 {/eq} and {eq}\displaystyle y(t) = t^3. {/eq} Suppose that {eq}\displaystyle \frac{\partial g}{\partial x} = 5 {/eq} and {eq}\displaystyle \frac{\partial g}{\partial y} = 3 {/eq} at {eq}\displaystyle (0, 1). {/eq} Use the Chain Rule to find {eq}\displaystyle \frac{dg}{dt} {/eq} when {eq}\displaystyle t = 1. {/eq}

## Total derivative:

If 'g' is any function made of more than two variables, also those variable are expressed in the form of any new independent variable i.e. parameter 't'. the total derivative of the function 'g' is given by the following formula:

{eq}\frac {dg}{dt} = \frac { \partial g}{ \partial x } \left ( \frac {dx}{dt} \right ) + \frac { \partial g}{ \partial y } \left ( \frac {dy}{dt} \right ) \\ {/eq}

When the function that has to be differentiated consist of two independent variables, then it has to be differentiated partially i.e. differentiating with respect to one independent variable only, taking other independent variable as constant

Gien:

{eq}\displaystyle x(t) = t - t^3 {/eq}

{eq}\displaystyle y(t) = t^3 \\ {/eq}

On calculating total derivative of 'g' with respect to 't', we get:

{eq}\displaystyle { \frac {dg}{dt} = \frac { \partial g}{ \partial x } \left ( \frac {dx}{dt} \right ) + \frac { \partial g}{ \partial y } \left ( \frac {dy}{dt} \right ) \\ = \frac { \partial g}{ \partial x } \left ( 1-3t^2 \right ) + \frac { \partial g}{ \partial y } \left ( 3t^2 \right ) \\ } {/eq}

On putting the values of {eq}\frac { \partial g}{ \partial x } \ and \ \frac { \partial g}{ \partial y } {/eq}, we get:

{eq}= 5 \left ( 1-3t^2 \right ) + 3 \left ( 3t^2 \right ) \\ {/eq}

On putting the value of 't' in the above equation, we get:

{eq}= 5 \left ( 1-3(1)^2 \right ) + 3 \left ( 3(1)^2 \right ) \\ = 5 \left ( 1-3(1) \right ) + 3 \left ( 3(1) \right ) \\ = 5 \left ( 1-3 \right ) + 3 \left ( 3 \right ) \\ = 5 ( -2) + 9 \\ = ( -10) + 9 \\ =-1 \\ {/eq}