Finding first partial derivatives of a function?
$begingroup$
So I have this function $f(x,y,z)=xsin(y-z).$
And I want to find :
the derivative with respect to $x$
the derivative with respect to $y$
the derivative with respect to $z .$
I think I should use the multivariable chain rule, but since I have dyscalculia I am not sure how to proceed adeguately .
I know that for the two variable chain rule,
$$text{if} quad f(x)=h(g(x)) quad text{then}quad f'=h'(g(x))cdot g'(x).$$ This formula for me is clear and I understand it well enough.
I have seen some formulas with three variables and they all seem very confusing for me so my question is,
could you please help me with a formula to find the derivative of a function composed of three variables with the same terms that i used in the equation above? meaning a formula with $h,g,$ possibly and variable, maybe $i?$ Possibly followed with partial derivatives formula related to a $3$ variable composite function
It would be really important for me if the formula could be in the same style of
$f(x)=h(g(x))$ then $f'=h'(g(x))·g'(x).;$ (so a formula for composite function )
Thanks !!
functions derivatives
edited Dec 6 '18 at 23:50
user376343
3,2982825
asked Dec 6 '18 at 17:16
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
So I have this function $f(x,y,z)=xsin(y-z).$
And I want to find :
the derivative with respect to $x$
the derivative with respect to $y$
the derivative with respect to $z .$
I think I should use the multivariable chain rule, but since I have dyscalculia I am not sure how to proceed adeguately .
I know that for the two variable chain rule,
$$text{if} quad f(x)=h(g(x)) quad text{then}quad f'=h'(g(x))cdot g'(x).$$ This formula for me is clear and I understand it well enough.
I have seen some formulas with three variables and they all seem very confusing for me so my question is,
could you please help me with a formula to find the derivative of a function composed of three variables with the same terms that i used in the equation above? meaning a formula with $h,g,$ possibly and variable, maybe $i?$ Possibly followed with partial derivatives formula related to a $3$ variable composite function
It would be really important for me if the formula could be in the same style of
$f(x)=h(g(x))$ then $f'=h'(g(x))·g'(x).;$ (so a formula for composite function )
Thanks !!
functions derivatives
edited Dec 6 '18 at 23:50
user376343
3,2982825
asked Dec 6 '18 at 17:16
BM97BM97
758
$endgroup$
3
$begingroup$
You don't need the chain rule for this problem. To take the partial derivative with respect to one of the variables, you treat the other two variables as if they were constants. All three partials are very simple in this example.
$endgroup$
– saulspatz
Dec 6 '18 at 17:30
$begingroup$
Thanks!!!!!!!!!!
$endgroup$
– BM97
Dec 6 '18 at 18:41
add a comment |
$begingroup$
So I have this function $f(x,y,z)=xsin(y-z).$
And I want to find :
the derivative with respect to $x$
the derivative with respect to $y$
the derivative with respect to $z .$
I think I should use the multivariable chain rule, but since I have dyscalculia I am not sure how to proceed adeguately .
I know that for the two variable chain rule,
$$text{if} quad f(x)=h(g(x)) quad text{then}quad f'=h'(g(x))cdot g'(x).$$ This formula for me is clear and I understand it well enough.
I have seen some formulas with three variables and they all seem very confusing for me so my question is,
could you please help me with a formula to find the derivative of a function composed of three variables with the same terms that i used in the equation above? meaning a formula with $h,g,$ possibly and variable, maybe $i?$ Possibly followed with partial derivatives formula related to a $3$ variable composite function
It would be really important for me if the formula could be in the same style of
$f(x)=h(g(x))$ then $f'=h'(g(x))·g'(x).;$ (so a formula for composite function )
Thanks !!
functions derivatives
edited Dec 6 '18 at 23:50
user376343
3,2982825
asked Dec 6 '18 at 17:16
BM97BM97
758
$endgroup$
So I have this function $f(x,y,z)=xsin(y-z).$
And I want to find :
the derivative with respect to $x$
the derivative with respect to $y$
the derivative with respect to $z .$
I think I should use the multivariable chain rule, but since I have dyscalculia I am not sure how to proceed adeguately .
I know that for the two variable chain rule,
$$text{if} quad f(x)=h(g(x)) quad text{then}quad f'=h'(g(x))cdot g'(x).$$ This formula for me is clear and I understand it well enough.
I have seen some formulas with three variables and they all seem very confusing for me so my question is,
could you please help me with a formula to find the derivative of a function composed of three variables with the same terms that i used in the equation above? meaning a formula with $h,g,$ possibly and variable, maybe $i?$ Possibly followed with partial derivatives formula related to a $3$ variable composite function
It would be really important for me if the formula could be in the same style of
$f(x)=h(g(x))$ then $f'=h'(g(x))·g'(x).;$ (so a formula for composite function )
Thanks !!
functions derivatives
functions derivatives
edited Dec 6 '18 at 23:50
user376343
3,2982825
asked Dec 6 '18 at 17:16
BM97BM97
758
edited Dec 6 '18 at 23:50
user376343
3,2982825
asked Dec 6 '18 at 17:16
BM97BM97
758
edited Dec 6 '18 at 23:50
user376343
3,2982825
edited Dec 6 '18 at 23:50
user376343
3,2982825
edited Dec 6 '18 at 23:50
user376343
3,2982825
3,2982825
asked Dec 6 '18 at 17:16
BM97BM97
758
asked Dec 6 '18 at 17:16
BM97BM97
758
asked Dec 6 '18 at 17:16
BM97BM97
758
758
3
$begingroup$
You don't need the chain rule for this problem. To take the partial derivative with respect to one of the variables, you treat the other two variables as if they were constants. All three partials are very simple in this example.
$endgroup$
– saulspatz
Dec 6 '18 at 17:30
$begingroup$
Thanks!!!!!!!!!!
$endgroup$
– BM97
Dec 6 '18 at 18:41
add a comment |
3
$begingroup$
You don't need the chain rule for this problem. To take the partial derivative with respect to one of the variables, you treat the other two variables as if they were constants. All three partials are very simple in this example.
$endgroup$
– saulspatz
Dec 6 '18 at 17:30
$begingroup$
Thanks!!!!!!!!!!
$endgroup$
– BM97
Dec 6 '18 at 18:41
3
3
$begingroup$
You don't need the chain rule for this problem. To take the partial derivative with respect to one of the variables, you treat the other two variables as if they were constants. All three partials are very simple in this example.
$endgroup$
– saulspatz
Dec 6 '18 at 17:30
$begingroup$
You don't need the chain rule for this problem. To take the partial derivative with respect to one of the variables, you treat the other two variables as if they were constants. All three partials are very simple in this example.
$endgroup$
– saulspatz
Dec 6 '18 at 17:30
$begingroup$
Thanks!!!!!!!!!!
$endgroup$
– BM97
Dec 6 '18 at 18:41
$begingroup$
Thanks!!!!!!!!!!
$endgroup$
– BM97
Dec 6 '18 at 18:41
add a comment |
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$begingroup$
You don't need the chain rule for this problem. To take the partial derivative with respect to one of the variables, you treat the other two variables as if they were constants. All three partials are very simple in this example.
$endgroup$
– saulspatz
Dec 6 '18 at 17:30
$begingroup$
Thanks!!!!!!!!!!
$endgroup$
– BM97
Dec 6 '18 at 18:41